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课后答案网：www.hackshp.cn11˜Ÿ4•Ø1˜Ü©4•ÐØ1˜ÙCþ†¼ê§1.¼êVg1.)eØ ª§¿xÑx‰Œµ1(1)−20)µ53(1)x<−½x>−22bb--3-2-10x(2)10ž§x<1½x>1+¶acc-011xwww.hackshp.cn1+a1a<0ž§1+||x|−|y||(2)|x1+x2+x3+···+xn|6|x1|+|x2|+···+|xn|(3)|x+x1+···+xn|>|x|−(|x1|+···+|xn|)y²µ22(1)Ï|x||y|>xy§K(x−y)>(|x|−|y|)§u´|x−y|>||x|−|y||(2)^êÆ8B{y².(i)n=2ž§d|x1+x2|6|x1|+|x2|§(ؤá.(ii)bn=kž(ؤá§=k|x1+x2+x3+···+xk|6|x1|+|x2|+···+|xk|.Kn=k+1ž§|x1+x2+x3+···+xk+1|6|x1+x2+x3+···+xk|+|xk+1|6|x1|+|x2|+···+|xk|+|xk+1|nþŒ•§é˜ƒg,ên§|x1+x2+x3+···+xn|6|x1|+|x2|+···+|xn|þ¤á.(3)|x+x1+···+xn|>|x|−|x1+x2+x3+···+xn|>|x|−(|x1|+···+|xn|)3.)eýéŠØ ª§¿xÑx‰Œµ(1)|x|>|x+1|1(2)2<<4|x|(3)|x|>A(4)|x−a|<η,η•~ê§η>0x−2x−2(5)>x+1x+1课后答案网1(6)2<<3|x+2|)µ1(1)x<−2www.hackshp.cnb--10x1111(2)−0ž§x<−A½x>Aee--A0AxA<0ž§x∈R(4)a−η0=(x+1)(x−2)>0§Kx>2½x6−1.7.f(x)=(|x|+x)(1−x)§¦÷veˆªxŠµ(1)f(0)=0课后答案网(2)f(x)<0)µ(1)‡f(x)=0§K|x|+x=0½1−x=0§=x60½x=1(2)Ï|x|+x>0§K‡f(x)<0§•‡1−x<0=Œ§=x>18.ã1-5L«>³|V!½>{R0Úwww.hackshp.cnŒC>{R|¤>´.3˜ãØ•žmS§A,Bü:m>ØVŒ±w¤˜‡~þ.¦Ñ>6IÚŒC>{R¼êª.)µd®•9ÔnÆ•£§V=I(R0+R).9.3˜‡Î/NìS?,«M—§TÎ/Nì.Œ»´a§p•h§?M—pÝ´x£ã1-6¤.TM—NÈVÚxƒm¼ê"XV=V(x)§¿Ñ§½Â•ÚŠ•.22)µd®•§V=πax§§½Â••[0,h]§Š••[1,πah]o10.,/Y± ¡È´˜‡F/§Xã1-7§.°2’§>–•45§CDL«Y¡§¦ ¡ABCD¡ÈS†Y h¼ê"X.)µd®•9ã§S=h(h+2).11.k˜ •H¶³§X^Œ»•Ròűz¦¨ωlÝ„Ýl¶³SåL•Ô§¦•Ô.¡†/¡ålsÚžmt¼ê"X£ã1-8¤.H)µd®•9ã§s=H−ωRtt∈0,ωt211+x,x<012.y=f(x)=§¦f(−2),f(−1),f(0),f(1)Úf.x−1,x>0211)µd®•§f(−2)=5,f(−1)=2,f(0)=−1,f(1)=0,f=−.22若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn50,06t<10213.x(t)=1+t,106t620§¦x(0),x(5),x(10),x(15),x(20),x(25),x(30)§¿xÑù‡¼êã/.t−10,20fé22u¤kx1,x2(x16=x2)¤á£…äkþãA5¼ê‰à¼ê¤.y²µ3•‚þ?ü:A(x1,f课后答案网(x1)),B(x2,f(x2))§ëAB§Ù¥:C(xC,yC)§Kf(x1)+f(x2)=2yC,x1+x2=2xCx1+x2x1+x2q•‚þxD=¤é:p‹I•yD=f§KxC=xD22f(x1)+f(x2)q•‚y=f(x)þ?˜^uÑpu§¤•l…x1,x2•u†l:§KyC>yD=>2x1+x2féu¤kx1,x2(x16=x2)¤á.2www.hackshp.cny6 B C f(x) A-0x1xDx2x19.y²eˆ¼ê3¤««mS´üNO¼êµ2(1)y=x(06x<+∞)ππ(2)y=sinx−6x622y²µ若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn721,0)Só¼ê§g1(x),g2(x)•½Â3(−a,a)(a>0)SÛ¼ê§F1(x)=f1(x)f2(x),F2(x)=g1(x)g2(x),F3(x)=f1(x)f2(x)Kf1(−x)=f1(x),f2(−x)=f2(x),g1(x)=−g1(x),g2(−x)=−g2(x)§u´F1(−x)=f1(−x)f2(−x)=f1(x)f2(x)=F1(x)F2(−x)=g1(−x)g2(−x)=(−g1(x))(−g2(x))=g1(x)g2(x)=F2(x)F3(−x)=f1(−x)g1(−x)=f1(x)(−g1(x))=−f1(x)g1(x)=−F3(x)lF1(x)´ó¼ê¶F2(x)´ó¼ê¶F3(x)´Û¼ê.23.f(x)•½Â3(−∞,+∞)S?Û¼ê§y²F1(x)≡f(x)+f(−x)´ó¼ê§F2(x)≡f(x)−f(−x)´Û¼ê.ÑéAue¼êF1(x),F2(x)µx(1)y=an(2)y=(1+x)y²µÏF1(−x)=f(−x)+f(x)=F1(x)§KF1(x)=f(x)+f(−x)´ó¼êqF2(−x)=f(−x)−f(x)=−F2(x)§KF2(x)=f(x)−f(−x)´Û¼ê.x−xx−x(1)F1(x)=f(x)+f(−x)=a+a,F2(x)=f(x)−f(−x)=a−annnn(2)F1(x)=f(x)+f(−x)=(1+x)+(1−x),F2(x)=f(x)−f(−x)=(1+x)−(1−x)24.`²e¼ê=´±Ï¼ê§¿¦•±Ïµ2(1)y=sinx2(2)y=sinx1(3)y=sinx+sin2x课后答案网2π(4)y=cosx4(5)y=|sinx|+|cosx|√(6)y=tanx(7)y=x−[x]www.hackshp.cn(8)y=sinnπx)µ2112π(1)Ïy=sinx=−cos2x§KT==π2222(2)by=sinx•˜±Ï¼ê…T=ω>022â±Ï¼ê½Â§é?Ûx∈√(−∞,+∞)§ksin(x+ω)=sinx§AOéx=0•AT¤á§22+Ksinω=0√§u´√ω=kπ,ω=kπ(k√∈Z)√222+qéx=2ω=2kπ•¤á§sin(2ω+ω)=sinω=0§K(2+1)kπ=nπ(n∈Z)§u√2k+´(2+1)=(k,n∈Z)n√2√−k+2q(2+1)=3+22∈Q§∈Q§Kbؤá§=¼êy=sinxØ´±Ï¼ê.n11(3)Ïy1=sinxT=2π¶y2=sin2xT=π§Ky=sinx+sin2xT=2π.222π(4)T=π=84若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn8πππ(5)Ïf(x)=|sinx|+|cosx|,fx+=sinx++cosx+=|cosx|+|sinx|=f(x)222πⲧ•y=|sinx|+|cosx|T=.2√(6)Ïf(x)=tanxT=π§Ky=tanxT=π.(7)Ïy=x−[x]=(x)§Ky=x−[x]T=1.2π2(8)T==nπn课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn9§2.EܼêÚ‡¼ê1.e¼êUĤEܼêy=f(ϕ(x))§XJU¤K•ÑdEܼê½Â•ÚŠ•µu2(1)y=f(u)=2,u=ϕ(x)=x2(2)y=f(u)=lnu,u=ϕ(x)=1−x231,x•knêž(3)y=f(u)=u+u,u=ϕ(x)=−1,x•Ãnêž(4)y=f(u)=2§½Â••U1§u=ϕ(x)§½Â••X§Š••U2√(5)y=f(u)=u,u=ϕ(x)=cosx)µu2(1)Ïy=f(u)=2½Â••(−∞,+∞)§u=ϕ(x)=xŠ••[0,+∞)x2Kd¼êU¤Eܼêy=2§§½Â••(−∞,+∞)§Š••[1,+∞)2(2)Ïy=f(u)=lnu½Â••(0,+∞)§u=ϕ(x)=1−xŠ••(−∞,1]2Kd¼êU¤Eܼêy=ln(1−x)§§½Â••(−1,1)§Š••(−∞,0]23(3)Ïy=f(u)=u+u½Â••(−∞,+∞)§1,x•knêžu=ϕ(x)=Š••{−1,1}−1,x•Ãnêž2,x•knêžKd¼êU¤Eܼêy=§§½Â••(−∞,+∞)§Š••{0,2}0,x•Ãnêž(4)Ïy=f(u)=2½Â••U1§u=ϕ(x)Š••U2TU1U26=φž§d¼êU¤Eܼêy=2§§½Â•ÀäN¼ê½§Š••{2}¶TU1U2=φž§d¼êØU¤Eܼê√(5)Ïy=f(u)=u½Â••[0,+∞)§u=ϕ(x)=cosxŠ••[−1,1]√hππiKd¼êU¤Eܼêy=cosx§§½Â••2kπ−,2kπ+(k=0,±1,±2,···)§Š•22•[0,1]22.f(x)=ax+bx+c§y²f(x+3)−3f(x+2)+3f(x+1)−f(x)≡0y²µd®•§222f(x+3)−3f(x+2)+3f(x+1)−f(x)=a(x+3)+b(x+3)+c−3[a(x+2)+b(x+2)+c]+3[a(x+1)+22222b(x+1)+c]−(ax+bx+c)=a[(x+3)−x]+b(x+3−x)−3a[(x+2)−(x+1)]−3b[x+2−(x+1)]=6ax+9a+3b−3a(2x+3)−3b课后答案网≡0c23.(1)y=f(x)=a+bx+§¦fxx2−x(2)y=f(x)=xln(1+x)§¦f(e)√(3)y=f(x)=1+x+x2§¦f(x2)9f(−x2)1(4)y=f(t)=√§¦fwww.hackshp.cn(atanx)a2+x2)µ2c22bc2bcxcx+2ax+4b(1)Ïy=f(x)=a+bx+§Kf=a++=a++=xxx2x22xxx2−x−x2−xln(e+1)−x(2)Ïy=f(x)=xln(1+x)§Kf(e)=(e)ln(1+e)=e2x√√√(3)Ïy=f(x)=1+x+x2§Kf(x2)=1+x2+x4,f(−x2)=1−x2+x41111(4)Ïy=f(t)=√§Kf(atanx)=p=√=a2+x2a2+(atanx)2a2sec2x|asecx|2x4.ef(x)=x,ϕ(x)=2§¦f(ϕ(x))9ϕ(f(x)).2xx22xxx2)µÏf(x)=x,ϕ(x)=2§Kf(ϕ(x))=(2)=2=4,ϕ(f(x))=23225.eϕ(x)=x+1§¦ϕ(x),(ϕ(x))9ϕ(ϕ(x)).3)µÏϕ(x)=x+1§K22362326333963ϕ(x)=(x)+1=x+1,(ϕ(x))=(x+1)=x+2x+1,ϕ(ϕ(x))=(x+1)+1=x+3x+3x+2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn10116.f(x)=§¦f(f(x)),f(f(f(x))),f.1−xf(x)1)µÏf(x)=§K1−x1x−11111f(f(x))==,f(f(f(x)))===x,f==1x1x−1f(x)1−(1−x)1−1−1−1−x1x1−1−x1x7.¦e¼ê‡¼ê9‡¼ê½Â•µ2(1)y=x(−∞0(a)f(x)=2−x,x<02(b)f(x)=x(2)òÿ ¼ê•µ(a)f(x)=sinx(b)f(x)=sin|x|3.‰e¼êã/µ(1)y=sgncosxhix(2)y=[x]−22)µybb6bb课后答案网1qqqq--π0πxbbb-1(1)y6cbwww.hackshp.cn1bbbbb--3-2-10123x(2)4.Š¼êy=(x)ã/.)µybbbc6bbb1 - -3-2-10123x5.Š¼êy=[x]−xã/.)µy6--3-2-10123x@@@@@@@@@@b@@b@@b@@c@@b@@b@@b-1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn126.˜‡¼ê´^eã•{û½µ3z˜‡«mn6x0)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn13y-π6bπ−b-kk--π+b-π+2b0π−2b3π−2bxk2k2k2ky6--2π0π2πx(3)9.e®•¼êy=f(x)ã/§Š¼êy1=|f(x)|,y2=f(−x),y3=−f(−x)ã/§¿`²y1,y2,y3ã/†yã/"X.)µy=f(x)ã/Xeµy6-0xKy1ã/•µy课后答案网6-0xKy2ã/•µwww.hackshp.cny6-0xKy3ã/•µy6-0xy1ã/f(x)<0ž†yã/"ux¶é¡§f(x)>0ž†yã/˜¶y2ã/†yã/"uy¶é¡§y3ã/†yã/"u:顧若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn14110.e®•f(x),g(x)ã/§ÁŠ¼êy={f(x)+g(x)+|f(x)−g(x)|}ã/§¿`²yã/†f(x),g(x)ã2/"X.)µy=max{f(x),g(x)}y-π6f(x)g(x)-0x11.éu½Â3[0,π]þ¼êy=x§kr§òÿ[0,2π]¦§"ux=π•é¡§, 2r®òÿ[0,2π]þ¼êòÿ‡¢¶þ¦¼ê•±2π•±Ï¼ê.x,x∈[0,π]2π−x,x∈[π,2π])µ¤¦¼ê•:f(x)=x−2nπ,x∈[2nπ,(2n+1)π](n=±1,±2,···)2nπ−x,x∈[(2n−1)π,2nπ](n=0,−1,±2,···)hixx+π=π−2π2πyπ6 @ @ @@ @@ @@ @ @--2π−π0π2πx课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn151Ù4•†ëY§1.ê4•ÚáŒþ1.Ñeêco‘µ13(1)xn=sinn3nm(m−1)···(m−n+1)n(2)xn=xn!111(3)xn=√+√+···+√n2+1n2+2n2+n√xn+yn(4)x1=a>0,y1=b>0,xn+1=xnyn,yn+1=211(5)x2n=1++···+(n=1,2,3,···)2n1x2n+1=(n=1,2,···)n):1111(1)x1=sin1,x2=sin8,x3=sin27,x4=sin6436912m(m−1)2m(m−1)(m−2)3(2)x1=mx,x2=x,x3=x,26m(m−1)(m−2)(m−3)4x4=x24111111(3)x1=√,x2=√+√,x3=√+√+√,2561011121111x4=√+√+√+√17181920r√√a+b(4)x1=a,x2=ab,x3=ab,r√√2√84a+ba+课后答案网bx4=ab··22√√2a+b(a+b)y1=b,y2=,y3=,√√2√p424(a+b)ab2(a+b)y4=+4163www.hackshp.cn1(5)x2=1,x3=1,x4=,x5=222.U½Ây²±eê•Ã¡þµn+1(1)n2+1sinn(2)nnn+(−1)(3)n2−11(4)n!111n+11(5)−+−···+(−1)n2n3nn2nn(6)(−1)(0.999)1−n(7)+en若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn16−ne(8)n√√(9)n+1−n1+2+3+···+n(10)n3y²:n+1n+12n2n+12(1)é∀ε>0§dun2+1−0=n2+1Nž§−0<εo¤á§¤±→0(n→∞)εn2+1n2+1sinnsinn1sinn11(2)é∀ε>0§dun−0=n6n§‡¦n−0<姕‡n<ε=Œ"N=ε+1§sinnsinnKn>Nž§−0<εo¤á§¤±→0(n→∞)nnn+(−1)nn+(−1)nn+11n+(−1)n(3)é∀ε>0§du−0=<=§‡¦−0<姕n2−1n2−1n2−1n−1n2−111n+(−1)nn+(−1)n‡<ε=Œ"N=+1§Kn>Nž§−0<εo¤á§¤±→n−1εn2−1n2−10(n→∞)111111(4)é∀ε>0§du−0=<§‡¦−0<姕‡<ε=Œ"N=+1§Kn!n!nn!nε11n>Nž§−0<εo¤á§¤±→0(n→∞)n!n!111n+11(5)Sn=−+−···+(−1)n2n3nn2111n+11é∀ε>0§duSn=(1−+−···+(−1))n23n11n+11δn1111δn=1−+−···+(−1)§KSn=n=2k+1ž§k0<δn=1−(−)−(−)−···−23nn2345111111111(−)<1¶n=2kž§k0<δn=1−(−)−(−)−···−(−)−<1"2k2k+123452k−22k−12kδn111oƒ§k0<δn<1l|Sn−0|=Sn=<‡¦|Sn−0|<姕‡<ε=Œ"N=+1§nnnε1111Kn>Nž§|Sn−0|<εo¤á§¤±−+−···+(−1)6n+1→0(n→∞)n2n3nn2课后答案网n−n11−n12(6)é∀ε>0§dun>lnn§Ke>n§u´e<§l+e−0=+e−n<§‡nnnnnnn1nn¦|(−1)(0.999)−0|<姕‡(0.999)<ε=Œ"N=2500ln+1§Kn>Nž§|(−1)(0.999)−εnn0|<εo¤á§¤±(−1)(0.999)→0(n→∞)1−n111−n22(7)é∀ε>0§dun+e−0=n!Nž§+e−0<εo¤á§¤±+e→0(n→∞)nne−ne−n1e−n1(8)é∀ε>0§due−nNž§−0<εo¤á§¤±→0(n→∞)εnn√√11√√1(9)é∀ε>0§du|n+1−n−0|=√√<√§‡¦|n+1−n−0|<姕‡√<n+1+n2n2n1√√√√ε=Œ"N=+1§Kn>Nž§|n+1−n−0|<εo¤á§¤±n+1−n→0(n→4ε2∞)1+2+3+···+nn+12n111(10)é∀ε>0§dun3−0=2n2<2n2=n§‡¦n−0<姕‡n<ε=Œ"11+2+3+···+n1+2+3+···+nN=ε+1§Kn>Nž§n3−0<εo¤á§¤±n3→0(n→∞)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn173.Þ~`²e"uáþ½Â´†Øµ(1)é?¿ε>0§•3N§n>Nž§¤áxn<ε¶(2)é?¿ε>0§•3Õõ‡xn§¦|xn|<ε.):n+1(1)~Xµê{−1+(−1)}(½{−n})={0,−2,0,−2,···}(½{−1,−2,−3,···})÷vþã^‡§Ø´Ã¡þ¶111(2)~Xµê{1,,1,,···,1,,···}÷vþã^‡§Ø´Ã¡þ"23n4.U½Ây²µ23n+n3(1)lim=n→∞2n2−12nz}|{(2)lim(0.99···9)=1n→∞√n2+n(3)lim=1n→∞n111(4)xn=++···+→1(n→∞)1·22·3(n−1)·nn−1n•óê(5)limrn=1§d?rn=n+1nn→∞n•Ûên3n=3k(k=1,2,3,···)3n+1n=3k+1(6)limrn=1§d?rn=nn→∞1+n2+√n=3k+23−n+ny²µ3n2+n32n+34(n+1)13n2+n3(1)é∀ε>0§du−=<=(n>2)§‡¦−<2n2−124n2−24(n+1)(n−1)n−12n2−12113n2+n3姕‡<ε=Œ"N=max(+1,2),Kn>Nž§−<εo¤á§¤n−1ε2n2−1223n+n3±→(n→∞)2n2−12z}|n{课后答案网z}|n{11(2)é∀ε>0§du0.99···9−1=(0.1)n=§‡¦0.99···9−1<姕‡<ε=Œ"N=10n10nz}|n{z}|n{1lg+1,Kn>Nž§0.99···9−1<εo¤á§¤±0.99···9→1(n→∞)ε√√√n2+nwww.hackshp.cnn2+n−n11n2+n(3)é∀ε>0§dun−1=n=<√2<2n§‡¦n−1<姕√n+n+n√11n2+nn2+n‡<ε=Œ"N=+1,Kn>Nž§−1<εo¤á§¤±→1(n→2n2εnn∞)1111111(4)é∀ε>0§duxn=1−+−+···+−=1−§K|xn−1|=§‡¦|xn−1|<ε§223n−1nnn11•‡<ε=Œ"N=+1,Kn>Nž§|xn−1|<εo¤á§¤±xn→1(n→∞)nεn±1111(5)é∀ε>0§du|rn−1|=−1=§‡¦|rn−1|<姕‡<ε=Œ"N=+1,Knnnεn>Nž§|rn−1|<εo¤á§¤±rn→1(n→∞)√1n−2n−4(6)é∀ε>0§du|r3k−3|=0,|r3k+1−3|=,|r3k+2−3|=√=√√Nž§|rn−3|<εo¤á§¤±rn→3(n→∞)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn185.(1)U½Ây²§ean→a(n→∞)§Ké?¿g,êk§an+k→a(n→∞)(2)U½Ây²§ean→a(n→∞)§K|an|→|a|.q‡ƒ´Ä¤áº(3)e|an|→0§Á¯an→a´Ä˜½¤áº•Ÿoºy²µ++(1)duan→a(n→∞)§é∀ε>0,∃N∈Z§n>Nž§|an−a|<ε§Ké∀k∈Z,n+k>+Nž§|an+k−a|<ε§u´é∀ε>0,∃N∈Z§n+k>Nž§|an+k−a|<ε§lan+k→a(n→∞)4d(Ø`²µKêc¡k•‘§•ØK•ÙÂñ5"(2)(i)duan→a§é∀ε>0,∃N∈Z+§n>Nž§|an−a|<ε.q|a<|an|−|a|n−a|§u´é∀ε>0,∃N∈Z+§n>Nž§|a<ε¤á§=|an|−|a|n|→|a|(n→∞(ii)‡ƒØ˜½¤á"~µn(a)ؤáµan=(−1)§K|an|→1§anÃ4•¶1(b)¤áµan=§K|an|→0,an→0n(3)du|an|→0§é∀ε>0,∃N∈Z+§n>Nž§|a<ε§q|a|a§un|−0n−0|=n|−0+´é∀ε>0,∃N∈Z§n>Nž§|an−0|<ε¤á§=an→0(n→∞)"le|an|→0§Kan→0˜½¤á"6.U½Ây²§exn→a§…a>b§K•3N§n>Nž§¤áxn>b.+y²µduxn→a§é∀ε>0,∃N∈Z§n>Nž§|xn−0|<ε§=a−εb§+a−b>0§Kε=a−b>0§l∃N∈Z§n>Nž§kxn>a−ε=a−(a−b)=b.=•3N§n>Nž§¤áxn>b.7.e{xnyn}Âñ§UÄä½{xn},{yn}½Âñ.)µØU"nn~µxn=(−1),yn=(−1)(n=1,2,···),xnyn≡1(n=1,2,···)§K{xnyn}Âñ§{xn},{yn}þØÂñ"e{xnyn}Âñ§ØUä½{xn},{yn}½Âñ.8.|^4•5Ÿ9OŽy²µ111(1)lim++···+=0n→∞n2(n+1)2(2n)2课后答案网111(2)lim√+√+···+√=1n→∞n2+1n2+2n2+nnPnkkn(n−1)2n(3)|^(1+h)=Cnh=1+nh+h+···+hk=02y²µnwww.hackshp.cn(i)lim=0(a>1)n→∞an5n(ii)lim=0(e≈2.7)n→∞eny²µ+111n+1n+1111(1)é∀n∈Z§k06++···+6§…lim=0§Klim++···+=n2(n+1)2(2n)2n2n→∞n2n→∞n2(n+1)2(2n)20+n111nn(2)é∀n∈Z§k<√+√+···+√<=1…lim=1§n+1n2+1n2+2n2+nnn→∞n+1111Klim√+√+···+√=1n→∞n2+1n2+2n2+nnnnn(3)(i)a=1+h(h>0)§du0<==<=an(1+h)nn(n1)2nn(n−1)21+nh+h+···+hh222212n§q•½Š§→0(n→∞)§K→0.llim=0(n−1)h2h2n−1(n−1)h2n→∞an若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn195555nnnn(ii)e=1+h(h≈1.7)§du0<==<0§Áyµ√√(1)xn→appm+am−1mm−1(2)a0xn1xn+···+am−1xn+am→a0a+a1a+···+am−1a+ammm−1(Ù¥a0a+a1a+···+am−1a+am>0)y²µ√√√+(1)duxn→a>0§é∀ε>0,∃N∈Z§n>Nž§|xn−a|0,∃N∈Z§n>Nž§|xn−a|<ε§lxn+aa√√xn→a(n→∞)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn20mm−1mm−1(2)duxn→a(n→∞)§a0xn+a1xn+···+am−1xn+am→a0a+a1a+···+am−1a+am>0§Kâ(1)ppamm−1mm−10xn+a1xn+···+am−1xn+am→a0a+a1a+···+am−1a+am11.éê{xn}§ex2k→a(k→∞),x2k+1→a(k→∞)§y²µxn→a(n→∞)+y²µ∀ε>0§Ïx2k→a(n→∞)§∃K1∈Z§¦k>K1ž§|x2k−a|<ε¤á"+qÏx2k+1→a(n→∞)§∃K2∈Z§¦k>K2ž§|x2k+1−a|<ε¤á"N=max{2K1,2K2+1}§Kn>Nž§en•óê§n=2k>N>2K1,k>K1,|xn−a|=|x2k−a|<ε§en•Ûê§n=2k+1>N>2K2+1,k>K2,|xn−a|=|x2k+1−a|<ε§Ïdxn→a(n→∞)12.|^üNk.7k4•§y²limxn•3§¿¦Ñ§µn→∞√√(1)x1=2,···,xn=2xn−1x0xn(2)x0=1,x1=1+,···,xn+1=1+1+x01+xny²µ√√(1)w,x1-n→∞§l=2l§)ƒl=2§=limxn=2"n→∞xn−11x0(2)w,xn>1§k^‡•xn=1+=2−<2§{xn}k."qx1=1+=1+xn−11+xn−11+x013111+=>1=x0§bxn-n→∞n→∞√√1+xn−1121+51−5n→∞§l=2−§=l=1+l§)l1=,l2=£ØÜK¿§¤§√1+l221+5=limxn="n→∞2√xn+yn13.ex1=a>0,y1=b>0(a0,yn>0(n∈Z).xn+1=xnyn>√xn+ynyn+ynxnxn=xn,yn+1=<课后答案网=yn§Kxn0,y1=b>0§22a-n→∞§α=αβqn→∞n→∞d01,k•ê)an√(4)xnn=a(00§xn+1>xn§K{xn}•üNO.q10§xn+1>xn§K{xn}•üNO.q1,k•ê§xn=>0§K{xn}ke."q==1+→anxnaan1+xn+1(n→∞)<1§∃N∈Z§n>Nž§k<1§KlN+1‘m©Ñkxn+1N)§l{xn}•34•"(4)dulnx1√nynn=lna=yn,0y2>···>yn>···§qxn−yn•Ã¡þ§{xn−yn}k.§|xn−yn|6C(n=1,2,···)£Ù¥C•,~ꤧK−C6xn−yn6C=xn6yn+C6y1+C§u´{xn}kþ.§l{xn}•34•"qyn>xn−C>x1−C§u´{yn}ke.§l{yn}•34•§Klimxn−limyn=lim(xn−yn)=0§u´limxn=limyn.n→∞n→∞n→∞n→∞n→∞16.x•?¿‰½¢ê§qyn(x)=sinsin···sinx§y²{yn(x)}4••3§¿¦d4•.|{z}ny²µk06x6π§K06sinx6x§lkyn+1(x)=sinyn(x)6yn(x)§{yn(x)}´±0•e.üNeü¼ê§7k4•§Ké∀x0∈[0,π]§k06u0=limyn(x0)=sinlimfn−1(x0)=sinu0§n→∞n→∞Ku0=0§lé∀x∈[0,π],limyn(x)=0.n→∞ÓnŒyx∈[−π,0]ž½klimyn(x)=0.n→∞2d±Ï5Œ•limyn(x)=0n→∞x1+x2+···+x+n17.elimxn=a§Áyµlim=an→∞n→∞n+εx1+x2+···+xny²µdlimxn=a§é∀ε>0,∃N1∈Z§n>N1ž§k|xn−a|<§Kk−a=n→∞2n(x1−a)+(x2−a)+···+(xn−a)|x1−a|+|x2−a|+···+|xN−a|+|xN−a|+···+|xn−a|611+10,∃N2=∈Z§n>N2ž§kε|x1−a|+|x2−a|+···+|xN−a|ε1Nž§k−a<+=ε§n22x1+x2+···+x+n=klim=an→∞nwww.hackshp.cnx1+x2+···+xn5µelim=a;limxn•3"n→∞nn→∞n−1x1+x2+···+xn~µxn=(−1)(n=1,2,···)§Kw,lim=0§limxnØ•3"n→∞nn→∞a1bn+a2bn−1+···+anb118.y²µeliman=a,limbn=b§Klim=abn→∞n→∞n→∞ny²µa1bn+a2bn−1+···+anb1(1)a=0§ylim=0n→∞n+dlimbn=b§Kâ½n4(P38)§∃M>0§¦|bn|6M(n∈Z)n→∞+ε2(|a1|+···+|an|)Mdliman=0§Ké∀ε>0,∃N1∈Z§n>N1ž§k|an|<.N=max+1,N1§n→∞2Mεa1bn+a2bn−1+···+anb1a1bn+a2bn−1+···+aNbn−N+aNbn−N+···+anb1u´n>N(>N1)ž§k=11+11+11nn(n−Nεa1bn+a2bn−1+···+aNbn−NaNbn−N+···+anb1(|a1|+···+|aN|)M1)··M611+1+1+1161+2M0§‡¦|n|>G§•‡n>G=Œ.N=[G]§Kn>Nž§|n|>Go¤á§√{n}´Ã¡Œþ"(2)é∀G>0§du|n!|>n§‡¦|n!|>G§•‡n>G=Œ.N=[G]§Kn>Nž§|n!|>Go¤á§{n!}´Ã¡Œþ"课后答案网GG(3)é∀G>0§‡¦|lnn|>G§•‡n>e=Œ.N=[e]§Kn>Nž§|lnn|>Go¤á§{lnn}´Ã¡Œþ"n2+1n2nn2+1n(4)é∀G>0§du>=§‡¦>G§•‡>G=Œ.N=[3G]§Kn>2n+13n32n+13n2+1n2+1Nž§>Go¤á§www.hackshp.cn{}´Ã¡Œþ"2n+12n+1n2+1n2nn2+1n(5)é∀G>0§du>=§‡¦>G§•‡>G=Œ.N=[2G]§Kn>2n−12n22n−12n2+1n2+1Nž§>Go¤á§{}´Ã¡Œþ"2n−12n−1nnn1111(6)é∀G>0§dulim1+=e…1+üNO§K1+ln2+ln+···+ln1+=ln(n+1)>lnn§K‡¦1+++···+>n23n2n23nG1111G§•‡lnn>G=Œ.N=[e]§Kn>Nž§1+++···+>Go¤á§{1++23n211+···+}´Ã¡Œþ"3n120.y²µe{xn}´Ã¡þ§xn6=0(n=1,2,3,···)§K´Ã¡Œþ"xn+y²µdu{xn}´Ã¡þ§é∀ε>0,∃N∈Z§n>Nž§k|xn|<ε若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn23111qxn6=0(n=1,2,3,···)§•3…>xnxnε11qε´?¿§•´?¿§l´Ã¡Œþ"εxn21.y²µe{xn}•Ã¡Œþ§{yn}•k.Cþ§K{xn±yn}•Ã¡Œþ"¿ddOŽe4•µ2n(1)limsinn+√n→∞n2+1(2)lim(n−arctann)n→∞n111(3)limn+(−1)++···+n→∞1·33·5(2n−1)(2n+1)qµü‡Ã¡ŒþÚ4•NºÁ?؈«ŒUœ/"i)y²µdu{yn}•k.Cþ§7•3êM§¦|yn|6M§q{xn}•Ã¡Œþ§é∀G>M>+0,∃N∈Z§n>Nž§k|xn|>G§Kn>Nž§k|xn±yn|>|xn|−|yn|>G−M.dG?¿59G>M>0§Œ•G−M>0…G−M´?¿§l{xn±yn}•Ã¡Œþ"ii))µ22nn(1)dulim√=∞…|sinn|61§Klimsinn+√=∞n→∞n2+1n→∞n2+1π(2)dulimn=∞…|arctann|6§Klim(n−arctann)=∞n→∞2n→∞nn111(−1)11111(3)xn=(−1)++···+§Kxn=1−+−+···+−=1·33·5(2n−1)(2n+1)23352n−12n+1nnn(−1)1(−1)2n(−1)111−=·=§k<|xn|<.qdlimn=∞§l22n+122n+1132n→∞2+nn111limn+(−1)++···+=∞n→∞1·33·5(2n−1)(2n+1)iii))µ(1)xn=n→+∞,yn=2n→+∞;xn+yn=3n→+∞(2)xn=−n→−∞,yn=−2n→−∞;xn+yn=−3n→−∞(3)xn=−n→−∞,yn=2n→+∞;xn+yn=n→+∞(4)xn=n→+∞,yn=−2n→−∞课后答案网;xn+yn=−n→−∞(5)xn=n+a→+∞,yn=−n→−∞;xn+yn=a£~þ¤nn(6)xn=n+(−1)→+∞,yn=−n→+∞;xn+yn=(−1)Ã4•22.?ØáŒþÚáþÚ!!û4•œ/")µ(1)Ú!µÏyn→0(n→∞)§www.hackshp.cn{yn}k."qxn→∞(n→∞)§KdþK(اk{xn±yn}•Ã¡Œþ"1ynxn(2)ûµxn6=0,yn6=0ž§duxn→∞,yn→0(n→∞)§Kkyn·→0§=→0,→∞xnxnyn23.Þ~`²Ã¡ŒþÚáþ¦ÈŒUu)««œ/")µ11(1)xn=n→+∞,yn=→0(n→∞);xn·yn=→0(n→∞)n2n21(2)xn=n→+∞,yn=→0(n→∞);xn·yn=n→+∞(n→∞)na(3)xn=n→+∞,yn=→0(n→∞);xn·yn=a£~þ¤nn1n(4)xn=n(−1)→∞,yn=→0(n→∞);xn·yn=(−1)Ã4•k.n2(−1)n1n1+(−1)n(5)xn=nn→∞,yn=→0(n→∞);xn·yn=n·n(−1)=nÃ4•§Ã.£…Ø´Ãn¡Œþ¤若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn2424.exn→∞,yn→a6=0§y²xnyn→∞111y²µduxn→∞(n→∞)§→0(n→∞)¶qyn→a6=0(n→∞)§→(n→∞)§uxnyna11´·→0(n→∞)§lxnyn→∞(n→∞)xnyn25.exn→+∞,yn→−∞§y²xnyn→−∞.+y²µÏxn→+∞§Ké∀G1>0,∃N1∈Z§n>N1ž§kxn>G1¶qyn→−∞§Ké∀G2>+0,∃N2∈Z§n>N2ž§k−yn>G2>0.N=max(N1,N2)§Kn>Nž§k−xnyn>G1G2§=xnyn<−G1G2.dG1,G2?¿5§G1G2´?¿…G1G2>0§Kxnyn→−∞.x1+x2+···+xn26.exn→+∞§y²→+∞n+x1+x2+···+xny²µÏxn→+∞§Ké∀G>0,∃N1∈Z§n>N1ž§kxn>3G§u´=nx1+···+xN1+nxN+···+xnx1+···+xNn−N11+1>1+·3G§nnnx1+·+xN1|x1|+···+|xN1|N1·MM=max(|x1|,···,|xN1|)§K66§u´éþãG>0§nnn2N1·Mx1+···+xN1Gx1+···+xN1Gn−N1N2=§Kn>N2ž§k<§l>−"qlim=Gn2n2n→∞n1+n−N11n−N111§éuε=,∃N3∈Z§n>N3ž§k−1<§l>§N=max{N1,N2,N3}§2n2n2x1+x2+···+xnG3Gx1+x2+···+xnKn>Nž§k>−+=G§dd•→+∞(n→∞).n22n课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn25§2.¼ê4•1.^©Û½Ây²µx−31(1)lim=x→−1x2−92x−31(2)lim=x→3x2−96x−1(3)lim√=2x→1x−1(x−2)(x−1)(4)lim=0x→1x−3t(t−1)1(5)lim=t→1t2−12x−1(6)lim=1x→∞x+2x(7)lim=∞x→3x2−92x+x(8)lim=∞x→∞x+1y²µx−3111x+1(1)é∀ε>0§dux2−9−2=x+3−2=2x+6§Ïx→−1§Ø”|x+1|<1§K−20§K0<|x−(−1)|<δž§Òkx2−9−2<εo¤á§xlim→−1x2−9=2x−3111x−3(2)é∀ε>0§dux2−9−6=x+3−6=6x+18§Ïx→3§Ø”|x−3|<1§K20§课后答案网K0<|x−3|<δž§Òkx2−9−6<εo¤á§xlim→3x2−9=6x−1√√x−1(3)é∀ε>0§du√x−1−2=|x+1−2|=|x−1|=√x+1§Ïx→1§Ø”|x−1|<1§√K00§K0<|x−1|<δž§Òk√−2<εo¤á§www.hackshp.cnx−1x−1lim√=2x→1x−1(x−2)(x−1)1(4)é∀ε>0§dux−3−0=1+x−3(x−1)§Ïx→1§Ø”|x−1|<1§K0<1212(x−2)(x−1)x<2§l0<1+x−3<3§u´1+x−3<3|x−1|§‡¦x−3−0<姕23(x−2)(x−1)‡|x−1|<ε=Œ"δ=minε,1>0§K0<|x−1|<δž§Òk−0<εo32x−3(x−2)(x−1)¤á§lim=0x→1x−3t(t−1)1t1t−1(5)é∀ε>0§dut2−1−2=t+1−2=2t+2§Ït→1§Ø”|t−1|<1§K00§K0<|x−(−1)|<δž§Òkt2−1−2<εo¤á§tlim→1t2−1=2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn26x−133(6)é∀ε>0§du−1=§Ïx→∞§Ø”|x|>2§K|x+2|>|x|−2§u´"X=+2§K|x|>Xž§Ò|x|−2x+2|x|−2εεx−1x−1k−1<εo¤á§lim=1x+2x→∞x+2xx1(7)é∀G>0§du=§Ïx→3§Ø”|x−3|<1§K27x−3§‡¦x2−9>G§•‡7x−3>G=2xxŒ"δ=min,1>0§K0<|x−3|<δž§Òk>Go¤á§lim=∞7Gx2−9x→3x2−9x2+xx2+x(8)é∀G>0§du=|x|§Ïx→∞§X=G>0§K|x|>Xž§Òk>Go¤x+1x+12x+xá§lim=∞x→∞x+12.¦4•µ2x−1(1)limx→02x2−x−12x−1(2)limx→12x2−x−12x−1(3)limx→∞2x2−x−1(1+x)(1+2x)(1+3x)−1(4)limx→0x2t(t−1)(5)limt→1t2−1√2t−t(6)lim√t→1t−1√1+x−2(7)limx→3x−35(1+x)−(1+5x)(8)limx→0x2+x5nm(1+mx)−(1+nx)课后答案网(9)lim£m,n•g,ê¤x→0x22x−5+6(10)limx→3x2−8x+152x+3x(11)limx→∞x25x−7(12)lim√www.hackshp.cnx→∞2x+x)µ2x−1(1)lim=1x→02x2−x−12x−1(x−1)(x+1)x+12(2)lim=lim=lim=x→12x2−x−1x→1(2x+1)(x−1)x→12x+132x−11(3)lim=x→∞2x2−x−12(1+x)(1+2x)(1+3x)−12(4)lim=lim(6+11x+6x)=6x→0xx→022t(t−1)t1(5)lim=lim=t→1t2−1t→1t+12√√√√2t−tt(t−1)(t+t+1(6)lim√=lim√=t→1√t−1√t→1t−1limt(t+t+1)=3t→1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn27√1+x−211(7)lim=lim√=x→3x−3x→31+x+2452345(1+x)−(1+5x)10x+10x+5x+x(8)lim=lim=10x→0x2+x5x→0x2+x5nm(1+mx)−(1+nx)(9)lim=x→0x22222233333nnmm22(Cnm−Cmn)x+(Cnm−Cmn)x+···+mx−nx2222nm−mnlim2=Cnm−Cmn=x→0x22x−5+6(x−2)(x−3)x−21(10)lim=lim=lim=−x→3x2−8x+15x→3(x−3)(x−5)x→3x−522x+3x(11)lim=1x→∞x25x−75(12)lim√=x→∞2x+x2P(x)3.R(x)=Q(x)ª¥P(x)ÚQ(x)•xõ‘ª§¿…P(a)=Q(a)=0§¯limk=ŒUŠºx→a)µduP(x)ÚQ(x)•xõ‘ª…P(a)=Q(a)=0§mnP(x)KP(x)=(x−a)P1(x),Q(x)=(x−a)Q1(x)(P1(a)6=0,Q1(x)6=0)§u´limR(x)=lim=x→ax→aQ(x)m(x−a)P1(x)limx→a(x−a)nQ1(x)?صP1(a)(1)n=mž§limR(x)=x→aQ1(a)m−nP1(x)P1(a)(2)n>mž§lim(x−a)=∞…lim=6=0§limR(x)=∞x→ax→aQ1(x)Q1(a)x→am−nP1(x)P1(a)(3)n0§0<|x−x0|<δžkf(x)>g(x)§y²A>B.x→x0x→x0qe0<|x−x0|<δžf(x)>g(x)§´Ä˜½¤áA>By²µ(1)^‡y{"bA0§¦0<|x−x0|f(x)"ù†®•µ∃δ>0§0<|x−x0|<δž§kf(x)>g(x)gñ§bؤá§=A>B¤á"(2)ؘ½"~µ242(x+3x)242(i)¤á"f(x)=,g(x)=x+3xx,∃δ>0§0<|x|<δž§kf(x)>g(x)"x2qA=limf(x)=2,B=limg(x)=1§A>B¤á"x→x0x→x024x+3x242(ii)ؤá"f(x)=,g(x)=x+xx,∃δ>0§0<|x|<δž§kf(x)>g(x)"qA=x2limf(x)=1,B=limg(x)=1§kA=B"x→x0x→x06.e3:x0•Skg(x)6f(x)6h(x)§¿…g(x)Úh(x)3x04••3¿…Ñ uA§y²limf(x)=x→x0A.y²µXJé?Ûxn,xn→x0,xn6=x0§¿…ŒØ”bxn∈O(x0,δ)−{x0}§kg(xn)6f(xn)6h(xn)±9g(xn)→A,h(xn)→A(n→∞)§dê4•5Ÿµf(xn)→A(n→∞)§ùÒy²f(x)→A(x→x0).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn29f(x)A7.elimf(x)=A,limg(x)=B6=0§y²lim=.x→x0x→x0x→x0g(x)Bf(x)ABf(x)−Ag(x)Bf(x)−AB+AB−Ag(x)|B||f(x)−A|+|A||g(x)−B|y²µ•g(x)−B=Bg(x)=BG(x)6|B||g(x)|§dulimf(x)=A,limg(x)=B§é∀ε>0,∃δ1>0§0<|x−x0|<δ1ž§k|f(x)−A|<ε¶éþx→x0x→x0ãε>0,∃δ2>0§0<|x−x0|<δ2ž§k|g(x)−B|<ε222BBqâ¦{$ŽµlimBg(x)=B>§Kâ5Ÿ3§∃δ3>0§0<|x−x0|<δ3ž§kBg(x)>x→x022f(x)A(|A|+|B|)ε2(|A|+|B|)δ=min{δ1,δ2,δ3}§0<|x−x0|<δž§k−<=εg(x)BB2B22f(x)A2(|A|+|B|)f(x)Au´§é∀ε>0,∃δ>0§0<|x−x0|<δž§k−<2ε§llim=.g(x)BBx→x0g(x)B0x>18.(1)f(x)=1x=12x+2x<1¦f(x)3x=1†m4•"1xsinx>0(2)f(x)=x21+xx<0¦f(x)3x=0†m4•")µ2(1)limf(x)=lim(x+2)=3,limf(x)=0x→1−0x→1−0x→1+02(2)limf(x)=lim(1+x)=1,x→−0x→−01limf(x)=lim(xsin)=0x→+0x→+0x9.`²e¼ê3¤«:†m4•œ/µ100§y²•3X>0§¦x<−X¤áµ0§é‰½ε=>0,∃X>0§x<−Xž§k|f(x)−A|<§x→−∞22A3=0,∃X1>0§x>X1ž§k|f(x)−A|<ε…∃X2>0,M>0§x→+∞x>X2ž§k|f(x)|0,∃X3>0§x>X3ž§k|g(x)−B|<ε.x→+∞X=max{X1,X2,X3}§éþãε>0§x>Xž§k|f(x)g(x)−AB|=|f(x)g(x)−f(x)B+f(x)B−AB|6|f(x)||g(x)−B|+|B||f(x)−A|6Mε+|B|ε=(M+|B|)ε§=limf(x)g(x)=AB.x→+∞16.y²limf(x)=A¿‡^‡´µé?Ûêxn→+∞,f(xn)→A.x→+∞y²µ⇒dulimf(x)=A§é∀ε>0,∃X>0§x>Xž§k|f(x)−A|<ε.x→+∞+qxn→+∞(n→∞)§éþãX>0,∃N∈Z§n>Nž§kxn>X§l|f(xn)−A|<ε§u´limf(xn)=A.n→∞000⇐^‡y{"blimf(x)6=A§K∃ε0>0§é∀X>0§–k˜‡x§x>Xž§k|f(x)−A|>x→+∞ε0.课后答案网000AO/§X•1,2,3,···§Œx1,x2,x3,···§¦000000x1>1ž§k|f(x1)−A|>ε0¶x2>2ž§k|f(x2)−A|>ε0¶x3>3ž§k|f(x3)−A|>ε0¶···00l†>Œ±wÑxn→+∞(n→∞)§lm>wÑlimf(xn)6=A§†®•gñ§Kbؤá§n→∞limf(x)=Ax→+∞17.y²limf(x)=+∞¿‡^‡´µé?Ûêxn:xn>x,xn→x§kf(xn)→+∞.00x→x+00y²µwww.hackshp.cn⇒dulimf(x)=+∞§é∀G>0,∃δ>0§0G.0x→x+00+qxn>x,xn→x(n→∞)§éþãδ>0,∃N∈Z§n>Nž§k0000G§u´limf(xn)=+∞.n→∞00⇐^‡y{"blimf(x)6=+∞§K∃G0>0§é∀δ>0§–k˜‡x§0Œ±wÑxn>x0,xn→x0§lm>wÑlimf(x)6=+∞§†®•gñ§Kbؤá§x→x+00limf(x)=+∞x→x+0018.ÞÑÎÜe‡¦f(x)(1)f(+0)=0,f(−0)=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn32(2)f(+0)Ø•3§•š∞,f(−0)=0(3)f(+∞)=0,f(−∞)Ø•3(4)f(+∞)=f(−∞)=A£~ê¤(5)f(x+0)Úf(x−0)ÑØ•300(6)f(x+0)=+∞,f(x−0)=−∞00(7)f(x+0)=1,f(x−0)=+∞00(8)f(+∞)Ø•3§•š∞,f(−∞)=−∞)µ0x>0(1)f(x)=1x60(1sinx>0(2)f(x)=x0x60−x(3)f(x)=eAx+1(4)f(x)=x1(5)f(x)=sinx−x01(6)f(x)=x−x0−1x−x(7)f(x)=1+e0sinxx>0(8)f(x)=xx<0课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn33§3.ëY¼ê1.U½Ây²e¼ê3½Â•SëYµ√(1)y=x1(2)y=x(3)y=|x|1(4)y=sinxy²µ√√|x−x0||x−x0|(1)x0•(0,+∞)S?˜:§|x−x0|<√√6√x+x0x0√√√|x−x0|√é∀ε>0§δ=x0ε§|x−x0|<δž§k|x−x0|<√<ε§y=x3x0:ëY.x√0qdx3(0,+∞)¥?¿5§Ky=x3(0,+∞)SëY.0√√√2x=0ž§éþãε>0§δ=ε§00,xx>0§u´|−|=0<002202xxxx00|x−x|02x022xxx11|x−x|ex•(−∞,0)S?˜:§Ø”|x−x|<−0§Kx<0,xx>0§u´|−|=0<002202xxxx00|x−x|02x02Sx•(−∞,0)(0,+∞)S?˜:§0|x|x211|x−x|é∀ε>0§δ=min0,0ε>0§|x−x|<δž§k−=0>ε§y=220xxxx0013x:ëYx0S1Sqdx3(−∞,0)(0,+∞)S?¿5§y=3(−∞,0)(0,+∞)SëY.0课后答案网x(3)x•(−∞,+∞)S?˜:§||x|−|x||6|x−x|.000é∀ε>0§δ=ε>0§|x−x|<δž§k||x|−|x||6|x−x|<ε§y=|x|3x:ëY0000qdx3(−∞,+∞)S?¿5§y=|x|3(−∞,+∞)SëY.0xxx211(4)x•(0,+∞)S?˜:§Ø”|x−x|<0§Kx>0,xx>0§u´sin−sin=002202xx0x+xx−x|x−x||x−x|2sin0cos06www.hackshp.cn0<02xx2xxxxx200002xxx211ex•(−∞,0)S?˜:§Ø”|x−x|<−0§Kx<0,xx>0§u´sin−sin6002202xx0|x−x||x−x|0<0xx20x02Sx•(−∞,0)(0,+∞)S?˜:§0|x|x211|x−x|é∀ε>0§δ=min0,0ε>0§|x−x|<δž§ksin−sin60<ε§220xxxx001y=sin3x:ëYx0S1Sqdx3(−∞,0)(0,+∞)S?¿5§y=sin3(−∞,0)(0,+∞)SëY.0x2.|^ëY¼ê$Ž§¦e¼êëY‰Œµ(1)y=tanx若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn341(2)y=xn(3)y=secx+cscx1(4)y=√cosxln(1+x)(5)y=x2−2x[x]tanx(6)y=1+sinx)µsinxππ(1)Ïy=tanx=§Kcosx6=0ž§y=tanxëY§y=tanxëY‰Œ•−+kπ,+kπ(k∈cosx22Z).1S1(2)en>0§Ky=ëY‰Œ•(−∞,0)(0,+∞)¶en60§Ky=ëY§=§ëY‰Œxnxn•(−∞,+∞).11(3)ÏsecxëY‰Œ•k−π0ž§y=√ëY§y=√ëY‰Œ•−+2kπ,+2kπ.cosxcosx221ln(1+x)SS(5)Ïln(1+x)x>−1žëY§x6=0,x6=2žëY§y=ëY‰Œ•(−1,0)(0,2)(2,+∞).x2−2xx2−2x[x]tanx[x]sinx[x]tanx(6)Ïy==§Ksinx6=1,cosx6=0,x/∈Z/{0}ž§y=ëY§1+sinx(1+sinx)cosx1+sinx[x]tanxππy=ëY‰Œ•x∈kπ−,kπ+…x/∈Z/{0}(k∈Z).1+sinx223.ïÄe¼êëY5§¿xÑÙã/.2x−4,ex6=2(1)y=x−24,x=2sinx课后答案网,x6=0(2)y=x1,x=0sinx,x6=0(3)y==|x|1,x=0www.hackshp.cn(4)y=[x])µ2x−4(1)Ïlimy=lim=lim(x+2)=4§…x=2ž§y=4§¼ê3x=2ëYx→2x→2x−2x→22x−4x6=2ž§y==x+2w,ëY§x−22x−4,ex6=2y=x−23(−∞,+∞)SëY.4,x=2sinxsinxsinxsinx(2)x6=0ž§y==½y=−w,ëY"qlim=1=f(0)§¼ê3x=0ëxxxx→0xY§sinx,x6=0u´y=x3(−∞,+∞)SëY.1,x=0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn35sinxsinxsinx(3)Ïlimy=lim=1,limy=lim=−1§limyØ•3"qx>0ž§y==x→+0x→+0|x|x→−0x→−0|x|x→0|x|sinxsinxsinxS§x<0ž§y==−§w,ëY§d¼ê3Ø0ëY§=3(−∞,0)(0,+∞)Sx|x|xëY.(4)Ïlimy=lim[x]=k,limy=lim[x]=k−1(k∈Z)§KlimyØ•3§x=k(k∈x→k+0x→k+0x→k−0x→k−0x→kZ)•y=[x]mä:§3mä:?mëYk0,∃δ>0§|x−x|<δž§k|f(x)−f(x)|<ε000||f(x)|−|f(x)||6|f(x)−f(x)|<ε§=é∀ε>0,∃δ>0§|x−x|<δž§k|f(x)−f(x)|<ε§0000|f(x)|3x:ëY0qdx3Iþ?¿5§•|f(x)|3Iþ•ëY022Ó|f(x)−f(x)|=|f(x)−f(x)||f(x)+f(x)|=|f(x)−f(x)||f(x)−f(x)+2f(x)|60000002|f(x)−f(x)|(|f(x)−f(x)|+2f(x))<ε(ε+2f(x))§f(x)3x:ëY000002qdx3Iþ?¿5§•f(x)3Iþ•ëY02(2)‡L5§e|f(x)|½f(x)ëY§f(x)ؘ½ëY.1,x>02(i)ØëY"~µf(x)=§|f(x)|=1Úf(x)=1þ3(−∞,+∞)SëY§f(x)3x=−1,x<00:ØëY¶2(ii)ëY"~µf(x)=x,Kf(x)!|f(x)|!f(x)3(−∞,+∞)SþëY"5.(1)¼êf(x)x=xžëY§¼êg(x)x=xžØëY§¯d¼êÚ3x:´ÄëYº000(2)x=xž¼êf(x)Úg(x)öÑØëY§¯d¼êÚf(x)+g(x)3®•:x´Ä7•ØëYº00):(1)^‡y{"bf(x)+g(x)3x:ëY"0Ïf(x)x=xžëY§KdëY¼ê5Ÿ§g(x)=[f(x)+g(x)]−f(x)xžëY†®•gñ"00bؤá§=f(x)+g(x)3x:ëY.0(2)ؘ½"1,x>0−1,x>0(i)ëYµ~µf(x)=,g(x)=3x=0ÑØëY§f(x)+课后答案网−1,x<01,x<0g(x)=03x=0ëY.12(ii)ØëYµ~µf(x)=g(x)=3x=0ÑØëY§f(x)+g(x)=3x=0ØëY.xx6.(1)¼êf(x)3xëY§¼êg(x)3xØëY¶00(2)x=xž¼êf(x)Úg(x)öÑØëY§¯d¼ê¦Èf(x)g(x)3®•:x´Ä7ØëYº00):www.hackshp.cn(1)ؘ½"1,x>0(i)ëYµ~µf(x)=03x=0ëY§g(x)=3x=0ØëY§f(x)g(x)=03x=0,x<00ëY.11(ii)ØëYµ~µf(x)=x3x=0ëY§g(x)=3x=0ØëY§f(x)g(x)=3x=0ØëY.x2x(2)ؘ½"1,x>0−1,x>0(i)ëYµ~µf(x)=,g(x)=3x=0ÑØëY§f(x)g(x)=−1,x<01,x<0−13x=0ëY.11(ii)ØëYµ~µf(x)=g(x)=3x=0ÑØëY§f(x)g(x)=3x=0ØëY.xx27.ef(x)3[a,∞)ëY§¿…limf(x)•3§y²f(x)3[a,∞)k..x→∞y²µdulimf(x)•3§Ø”limf(x)=Ax→∞x→∞Kéε=1,∃X>0§x>Xž§k|f(x)−A|<ε=1¤á§l|f(x)|=|f(x)−A+A|6|f(x)−A|+若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn36|A|<1+|A|X1=max{X,a+1}§Kf(x)3(X1,∞)Sk.§…|f(x)|<|A|+1,x∈(X1,∞)qduf(x)3[a,X1]þëY§f(x)3[a,X1]þk.§Ù.•M>0§=∀x∈[a,X1]§k|f(x)|6MG=max{|A|+1,M}§K∀x∈[a,∞),f(x)6G§=f(x)3[a,∞)k..8.eé?˜ε>0§f(x)3[a+ε,b−ε]ëY§¯µ(1)f(x)´Ä(a,b)3ëYº(2)f(x)´Ä3[a,b]ëYº)µx−ab−x(1)?x∈(a,b)§ε=min0,0§Kx∈[a+ε,b−ε]0220Ïé?˜ε>0§f(x)3[a+ε,b−ε]ëY§f(x)3x:ëY0dx∈(a,b)?¿5§f(x)3(a,b)SëY.0(2)ؘ½ëY"(i)ØëY"~µf(x)3[0+ε,1−ε](ε>0)SëY§f(x)3[0,1]þØëY§3x=0:äm.(ii)ëY"~µf(x)3[1+ε,2−ε](ε>0)SëY§…f(x)3[1,2]þëY.9.ef(x)3x:ëY§¿…f(x)>0§y²•3xδ•O(x,δ)§x∈O(x,δ)ž§f(x)>c>0§c•,00000‡~ê.y²µduf(x)3x:ëY§…f(x)>0§Kf(x)>c>0000鉽ε=f(x)−c>0,∃δ>0§|x−x|<δž§k|f(x)−f(x)|<ε=f(x)−c§Kf(x)−[f(x)−000000c]6f(x)§=f(x)>c>0.10.y²eëY¼ê3kn:¼êŠ•0§Kd¼êð•0.y²µf(x)•¢¶þëY¼ê§x•?¿˜‡Ãn:.0dkn:3ê¶þÈ—5§Œ±Ãnê{xn}§¦xn→x(n→∞).0Ïf(x)3xëY§Kf(x)=limf(xn)=0§00n→∞dx:?¿5§f(x)3¤kÃn:¼êŠÑ•0.0qf(x)3kn:¼êŠ•0§Kd¼êð•0.111.ef(x)3[a,b]ëY§ð§U½Ây²3[a,b]ëY.f(x)1y²µduf(x)3[a,b]ëY§ð§Kf(x)3(a,b)ëY§f(x)>0§•3§x∈[a,b]f(x)x•(a,b)S?˜:§Ké∀ε>0,∃δ>0§|x−x|<δž§k|f(x)−f(x)|<ε.000qf(x)3[a,b]ëY§Kd4«më课后答案网Y¼ê5Ÿ2§Œf(x)3[a,b]þ•Š•m>0§=f(x)>m,x∈11|f(x)−f(x)|ε111[a,b]§u´−=0<§lim=§l3xëY.f(x)f(x)f(x)f(x)m2x→xf(x)f(x)f(x)00000dx3(a,b)S?¿5§f(x)3(a,b)SëY.011qf(a+0)=f(a)>0§K=§f(x)3[a,b)ëY¶f(a+0)f(a)11qf(b−0)=f(b)>0§K=§f(x)3[a,b]ëY.f(b−0)www.hackshp.cnf(b)12.ef(x)Úg(x)Ñ3[a,b]ëY§Áy²max(f(x),g(x))±9min(f(x),g(x))Ñ3[a,b]ëY.y²µduf(x)Úg(x)Ñ3[a,b]ëY§f(x)−g(x)Úf(x)+g(x)Ñ3[a,b]ëY.d14K(اk|f(x)−g(x)|3[a,b]ëY.1-ϕ(x)=max(f(x),g(x))=(f(x)+g(x)+|f(x)−g(x)|),21ψ(x)=min(f(x),g(x))=(f(x)+g(x)−|f(x)−g(x)|)§2ϕ(x),ψ(x)Ñ3[a,b]ëY.−c,ef(x)<−c13.ef(x)´ëY§y²é?Ûc>0§¼êg(x)=f(x),e|f(x)|6c´ëY.c,ef(x)>cy²µdug(x)=max(−c,min(f(x),c))qduf(x)ëY§…é?Ûc>0§ϕ(x)=cëY§ψ(x)=−cëY§KdþK(اmin(f(x),c)ëY§l2dþK(اg(x)ëY.14.ïÄe¼êˆ‡ØëY:5Ÿ£=•Û«ØëY:¤µx(1)y=(1+x)2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn371+x(2)y=1+x32x−1(3)y=x3−3x+2x(4)y=sinx21(5)y=cosx(6)y=[x]+[−x]1(7)y=lnx2x−x(8)y=|x|(x2−1)1p,x=(q>0,q,p•pŸê)(9)y=qq0,x•Ãnêx,|x|61(10)y=1,|x|>1(πxcos,|x|61(11)y=2|x−1|,|x|>1sinπx,x•knê(12)y=0,x•Ãnê)µx(1)Ïlim=−∞§x=−1•1aØëY:£Ã¡mä:¤.x→−1−0(1+x)21+x1(2)Ïlim=§y3x=−1:vk½Â§x=−1•Œ£ØëY:.x→−11+x332x−1(x−1)(x+1)(x−1)(x+1)(x−1)(x+1)(3)Ïy====§x3−3x+2(x−1)(x2+x+1)−3(x−1)(x−1)(x2+x−2)(x−1)2(x+2)qlimy=−∞,limy=−∞§x=−2,x=1•1aØëY:.x→1−0x→−2−0x(4)Ïlim=1y3x=0ý§x=0•Œ£ØëY:¶x→0sinxxqlim=∞§x=kπ(k∈Z,k6=0)•1aØëY:.x→kπsinx课后答案网k∈Z,k6=021(5)Ïlimcos3[0,1]m§•.4•§d4•Ø•3§u´x=0•1aØëY:.x→0x(6)Ïx→k+0ž§−x→−k−0§limy=lim([x]+[−x])=k+(−k−1)=−1¶x→k+0x→k+0qÏx→k−0ž§−x→−k+0§limy=lim([x]+[−x])=k−1+(−k)=−1(k∈Z)x→k−0x→k−0qx=kž§y=[x]+[−x]=[www.hackshp.cnk]+[−k]=0(k∈Z)§ê:þ•Œ£ØëY:.1(7)Ïlim=+∞§x=−1•1aØëY:¶x→1+0lnx1ÏlimØ•3§x=0•1aØëY:.x→−0lnxx(x−1)(8)y=|x|(x−1)(x+1)1Ïlimy=y3x=1ý§x=1•Œ£ØëY:¶x→12Ïlimy=1,limy=−1§x=0•1˜aØëY:£amä:¤¶x→+0x→−0Ïlimy=−∞§x=−1•1amä:.x→−1+0(9)Ïd¼ê´±1•±Ï¼ê§Œ3«m[0,1]?ا٧«mœ/†daq.0113[0,1]þ§©1•1knêkü‡µ,¶©1•2knêk˜‡µ¶1121213©1•3knêkü‡µ,¶©1•4knêkü‡µ,¶3344123415©1•5knêko‡µ,,,¶©1•6knêkü‡µ,¶···555566若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn38k(k−1)oƒ§©1؇Lkknê‡êl62+1+2+3+···+(k−1)=+2§=©1؇Lkk2nê•kk•‡"e¡5y§3?˜:x∈[0,1]§x→xž§y→0.001é∀ε>0§k=§3[0,1]þ§©1؇Lkknê•r1,r2,···,rl.εδ=minlimits16i6l|ri−x|§K0<|x−x|<δ§=x/∈{r1,r2,···,rn}§•Ò´x½ö•Ãnꧽö•k0011pp6,x•knêx=,q>kn꧅q>k+1>kž§Òk|y−0|=qk+1q.q0<ε,x•Ãnêlimy=0§u´µ?ÛÃn:Ñ´d¼êëY:§?Ûkn:Ñ´d¼êŒ£ØëY:.x→x0(10)Ïlimy=−1,limy=1§x=−1•1˜aØëY:.x→−1+0x→−1−0(11)Ïlimy=0,limy=2§x=−1•1˜aØëY:.x→−1+0x→−1−0(12)(i)x6=n,n∈Z§0kn:rn→x…rn>x§Klimf(rn)=sinπx6=0¶000rn→x+00Ãn:xn→x…xn>x§Klimf(xn)=0"00xn→x+00f(x+0)Ø•3§lx6=n(n∈Z)•¼ê1aØëY:.0(ii)x=n,n∈Z§0x•Ãnꞧ|f(x)−f(n)|=0¶εx•knꞧ|f(x)−f(n)|6π|x−n|§é∀ε>0,∃δ=>0§¦|x−n|<δž§k|f(x)−πf(n)|<ε§f(x)3x=n(n∈Z)ëY.15.x=0že¼êf(x)ý§Á½Âf(0)ꊧ¦f(x)3x=0ëYµ√1+x−1(1)f(x)=√31+x−1tan2x(2)f(x)=x1(3)f(x)=sinx·sinx1(4)f(x)=(1+x)x)µ课后答案网√p√3(1+x)2+31+x+11+x−13(1)Ïlimf(x)=lim√=lim√=§x→0x→031+x−1x→01+x+123f(0)=.2tan2x(2)Ïlimf(x)=lim=2§x→0x→0xwww.hackshp.cnf(0)=2.1(3)Ïlimf(x)=limsinx·sin=0§x→0x→0xf(0)=0.1(4)Ïlimf(x)=lim(1+x)x=e§x→0x→0f(0)=e.f(x1)+f(x2)+···+f(xn)16.ef(x)3[a,b]ëY§am.nf(x1)+f(x2)+···+f(xn)duf(x)3[x1,xn]⊂[a,b]þëY§d0Š½n•§7∃ξ∈[x1,xn]⊂[a,b]§¦f(ξ)=.n17.^˜—ëY½Âyµ若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn39√(1)f(x)=3x3[0,1]þ´˜—ëY¶(2)f(x)=sinx3(−∞,+∞)þ´˜—ëY¶2(3)f(x)=sinx3(−∞,+∞)þؘ—ëY.y²µ√√|x1−x2||x1−x2|(1)é?Ûx1,x2∈[0,1]§|3x1−3x2|=p√p=√√√√63x23x323(3x1+3x2)2+1(3x1−3x2)21+1x2+x244|x1−x2|√√§1(3x321−x2)41√3√33√3√3p3=|x1−x2|6|x1−x2|§½=|x1−x2|64|x1−x2|4ε3√√p>0§¦é∀x333é∀ε>0,∃δ=1,x2∈[0,1]§|x1−x2|<δž§ok|x1−x2|64|x1−x2|<4ε√lf(x)=3x3[0,1]þ´˜—ëY.x1+x2x1−x2x1−x2(2)é?Ûx1,x2∈(−∞,+∞)§|sinx1−sinx2|=2cossin62=|x1−x2|§222é∀ε>0,∃δ=ε>0§¦é∀x1,x2∈(−∞,+∞)§|x1−x2|<δž§ok|sinx1−sinx2|6|x1−x2|<εlf(x)=sinx3(−∞,+∞)þ´˜—ëY.pppp0π00π000ππ(3)ε0=1§é?Ûδ>0§xn=2nπ+2,xn=2nπ−2§|xn−xn|=|2nπ+2−2nπ−2|=πpp→0(n→∞)2nπ+π+2nπ−π22000n¿©Œž§˜½k|xn−xn|<δ§022002|sin(xn)−sin(xn)|=|1−(−1)|=2>1=ε02lf(x)=sinx3(−∞,+∞)þؘ—ëY.课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn40§4.áþÚáŒþ 1.¦eáþx→0ž Ú̇ܩµ36(1)x+x235(2)4x+6x−x√(3)x·sinxp√(4)x2+3x√√(5)1+x−1−x(6)tanx−sinx(7)ln(1+x))µ36x+x33(1)dulim=lim(1+x)=1§§´˜‡3 áþ§§Ì‡Ü©•x.x→0x3x→023534x+6x−x3x2(2)dulim=lim(1+x−)=1§§´˜‡2 áþ§§Ì‡Ü©•4x.x→04x2x→024√rx·sinxsinx(3)dulim=lim=1§§´˜‡1 áþ§§Ì‡Ü©•|x|.x→0|x|x→0xp√qx2+3x√516(4)dulim√=limx3+1=1§§´˜‡ áþ§§Ì‡Ü©•x.x→06xx→06√√1+x−1−x2x(5)dulim=lim√√=1§§´˜‡1 áþ§§Ì‡Ü©x→0xx→0x(1+x+1−x)•x.2tanx−sinxtanx−sinx2(1−cosx)x(6)dulim=lim2=lim=lim=1§§´˜‡3 áx→0x3x→0x3x→0cosx·x2x→0x223xþ§§Ì‡Ü©•.2ln(1+x)(7)dulim=1§§´˜‡1 áþ§§Ì‡Ü©•x.x→0x2.x→∞ž§¦eCþ ÚÌ课后答案网‡Ü©µ26(1)x+x245(2)4x+6x−xr321(3)xsinxqp√(4)1+1+xwww.hackshp.cn52x(5)x3−3x+1)µ26x+x6(1)dulim=1§§´˜‡6 áŒþ§§Ì‡Ü©•x.x→∞x62454x+6x−x5(2)dulim=1§§´˜‡5 áŒþ§§Ì‡Ü©•−x.x→∞−x5r321√xsinx3x1√(3)dulim√=lim√=1§§´˜‡ áŒþ§§Ì‡Ü©•3x.x→∞3xx→∞3x3qpvus√u111+1+xt14121(4)dulim√=lim++1=1§§´˜‡ áŒþ§§Ì‡x→∞8xx→∞xx8√Ü©•8x.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn4152x3x3−3x+1x2(5)dulim=lim=1§§´˜‡2 áŒþ§§Ì‡Ü©•2x.x→∞2x2x→∞x3−3x+13.Áyµ∆x→0žmnn(1)o(∆x)+o(∆x)=o(∆x)(m>n>0)mnm+n(2)o(∆x)o(∆x)=o(∆x)(m,n>0)(3)|f(x)|6M§Kf(x)o(∆x)=o(∆x)mm(4)∆x·o(1)=o(∆x)y²µmnmno(∆x)o(∆x)(1)du∆x→0§∆x→0,∆x→0§u´→0,→0∆xm∆xnmmnmmn∆xm−no(∆x)+o(∆x)o(∆x)∆xo(∆x)qm>n>0§=∆x→0§u´=·+→0§∆xn∆xn∆xm∆xn∆xnmnnlo(∆x)+o(∆x)=o(∆x)mnmno(∆x)o(∆x)(2)du∆x→0§∆x→0,∆x→0§u´→0,→0∆xm∆xnmnmno(∆x)o(∆x)o(∆x)o(∆x)u´=·→0§∆xm+n∆xm∆xnmnm+nlo(∆x)o(∆x)=o(∆x)o(∆x)f(x)o(∆x)o(∆x)(3)∆x→0§→0§q|f(x)|6M§f(x)k.§u´=f(x)→0§l∆x∆x∆xf(x)o(∆x)=o(∆x).mm∆x·o(1)∆xm(4)do(1)u´Ã¡þ§Ko(1)→0§u´=o(1)=o(1)→0§l∆x·o(1)=∆xm∆xmmo(∆x).课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn421Ü©4•YØ1nÙ"u¢êÄ ½n94«mþëY¼ê5Ÿy²§1."u¢êÄ ½n1.l½ÂÑuy²e(.•˜5.000y²µα,αÑ´ê8Ee(.§u´∀x∈E§Ñkx>α§=α´Ee.¶x>α§=α´Ee..000duα´Ee(.§´e.¥•Œö§lkα>α¶Ódα´Ee(.§kα>α.dd0•α=α.2.β=supE,β/∈E§ÁygE¥ŒÀê{xn}§Ù4••β¶qeβ∈E§Kœ/XÛºy²µ(1)duβ=supE,β/∈E§Kdþ(.½Â§(i)é∀x∈E§Ñkx<β¶(ii)é∀ε>0§–•3˜‡êx∈E§¦x>β−ε.001εn=§éz‡εnÑkxn∈E§¦β>xn>β−εn§=0<β−xn<εn§u´ŒÀ˜‡ên{xn}⊂E.qlim(β−εn)=β−limεn=β…β>limxn>lim(β−εn)=β§limxn=β.n→∞n→∞n→∞n→∞n→∞111(2)β∈Ež§·Kؘ½¤á"~µØ¤á"E=(1,,,···,,···),β=supE=1,1∈E.23n1q→0(n→∞)§KE¥?˜f4•þ•0§β∈Ež§·Kؤá"nπ2πnπ16n+4¤á"E=sin,sin,···,sin,···,β=supE=1,1∈E§xn=sinπ§Klimxn=8888n→∞1§β∈Ež§·K¤á"3.Þ~µ(1)kþ(.Ãe(.ê¶(2)¹kþ(.عke(.课后答案网ê¶(3)Q¹kþ(.q¹ke(.ê¶(4)Qعkþ(.§qعke(.ê§Ù¥þ!e(.Ñk•.)µ(1){xn}={−n},sup{xn}=−11(2){xn}={},sup{xn}=1∈{www.hackshp.cnxn},inf{xn}=0∈{/xn}nn(3){xn}={1+(−1)},sup{xn}=2∈{xn},inf{xn}=0∈{xn}11121n−1(4)E=1,,1+,,1+,···,,1+,supE=2∈/E,infE=0∈/E2233nn4.ÁyÂñê7kþ(.Úe(.§ªu+∞ê7ke(.§ªu−∞ê7kþ(..y²µ(1)éuˆ‘ð•~êê§w,þ!e(.þŒˆ.éuØð•~êê§Ï{xn}Âñ§={xn}k4•§Kd1Ù§1½n4§ê{xn}´k.ê.ld Ù½nn§ê{xn}kþ!e(.§=Âñê7kþ!e(..5µ„Œy²µþ!e(.β,α¥–k˜‡áu{xn}.¯¢þ§eα=β§Kα=β=xn,n=1,2,···nono(1)(2)eα6=β§…α/∈{xn}§KdSK2•§•3fxnkÂñu᧕•3fxnkÂñuβ§{xn}ØÂñ§ù†®•{xn}Âñgñ§α,β¥–k˜‡áu{xn}.+(2)Ï{xn}´ªu+∞ê§K∃N∈Z§n>Nž§ðkxn>x1§u´x1,x2,···,xN¥•ö§=•{xn}e(."若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn43+(3)Ï{xn}´ªu−∞ê§K∃N∈Z§n>Nž§ðkxn1,k=1,2,3,···;x2k+1>1…min{x2k}=1§inf{xn}=1£Œˆ¤.6.y²µüN~ke.ê7k4•.y²µdu{yn}ke.§{yn}7ke(..de(.½Âkµ(i)yn>α(n=1,2,3,···)¶(ii)é∀ε>0§–k˜‡y∈{yn}§¦y<α+ε.NNdu{yn}´üN~ê§n>Nž§kyn<α+ε§=n>Nž§k06yn−α<ε§u´yn→α(n→∞).lüN~ke.ê7k4•.7.Á©Û«m@½n^‡µeò4«mU•m«m§(JXÛºeò^‡[a1,b1]⊃[a2,b2]⊃···K½ò^‡bn−an→0K§(JNºÁÞ~`².)µ(1)3«m@½n¥§eò4«mU•m«m§=(i)(an+1,bn+1)⊂(an,bn)¶(ii)lim(bn−an)=0n→∞KŒ±y²{an},{bn}EÂñuÓ˜4•ξ§=liman=limbn=ξ§džξŒUŠ Øáuùmn→∞n→∞+«m§=ξ/∈(an,bn)(n∈Z)§½=ξŒUØ•(an,bn)ú:.1~µm«m(0,)§n11(i)0,⊂0,¶课后答案网n+1n11(ii)lim−0=lim=0¶n→∞nn→∞n11an=0→0(n→∞);bn=→0(n→∞)§Kξ=0∈/0,§=(Øؤá.nn(2)eò^‡[an+1,bn+1]⊂[an,bn]K§=•k^‡bn−an→0¤á§KØUy{an}†{bn}Âñ.www.hackshp.cn1111211~µ4«mn−,n+Ø´˜‡@˜‡.limn+−n−=lim=0§limn+†limn−nnn→∞nnn→∞nn→∞nn→∞nØÂñ.Ø•3ξ•{an},{bn}ú4•§=(Øؤá.(3)eò^‡bn−an→0K§=•k^‡[an+1,bn+1]⊂[an,bn]¤á.KŒ±y²{an},{bn}Âñ£†«m@½ny²˜¤§ØUylimbn=liman¤á§l[an,bn]ú:Ø•˜§$–Ñy˜‡ún→∞n→∞«m.1111+11~µ4«m1−,2+⊂1−,2+,n∈Z§…lim2+−1−=1.n+1n+1nnn→∞nn11+dliman=1,limbn=2§[1,2]⊂1−,2+,n∈Z§=(Øؤá.n→∞n→∞nn(1)(2)8.e{xn}Ã.n§…šoáŒþ§K7•3ü‡fxnk→∞,xnk→a(a•,k•ê).(1)y²µkyxnk´˜‡Ã¡Œþ.0+du{xn}Ã.§é?Û¢êM>0§–k˜‡n∈Z§¦|xn0|>M.(1)(1)M=1§K7•3n1§¦xn1>1¶M=2§K7•3n2§¦xn2>2¶···¶M=K§K7•若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn44(1)3nK>nK−1§¦xnK>K§···.no(1)+(1)(1)KŒ˜fxnk§é∀M∈Z§K=M§Kk>Kž§Òkxnk>M§klimxnk=∞.k→∞++d®•{xn}شáŒþ§Kd½Â§∃M>0§é∀N∈Z§–k˜‡m∈Z§m>Nž§0k|xm|m§¦|xm|6M000102N=m1§K–k˜‡m2>m1§¦|xm|6M§···20Xd?1e§KŒ˜mtµm10§3(a−ε0,a+ε0)k{xn}áõ‘§¤{xn}f§P•xn.no(2)(2)duxnk.§•3fxnk→b§w,a6=b.nono(1)(2)(1)(2)10.e3«m[a,b]¥ü‡êxn9xn÷vxn−xn→0(n→∞)§K3düê¥UéäkƒÓv(1)(1)Inkfn§¦oxnk→x0,xnnk→ox0(k→∞).nono(1)(1)(1)(1)y²µÏxn⊂[a,b]§Kxn•˜k.ê§Kd——5½n§xn7kÂñf§P•xnk§(1)…limxn=x0.nk→∞onono(2)(1)(2)3xn¥Ñ†xnkkƒÓvIfxnk.(1)(2)(1)(2)Ïxn−xn→0(n→∞)§Klimxnk−xnk=0§hk→∞i(2)(1)(1)(2)(1)(1)(2)u´limxnk=limxnk−xnk−xnk=limxnk−limxnk−xnk=x0−0=x0.k→∞k→∞k→∞k→∞11.|^…ÜÂñn?ØeêÂñ5µ2n(1)xn=a0+a1q+a2q+···+anq(|q|<1,|ak|6M)sin1sin2sinn(2)xn=1+++···+2222n11n+11(3)xn=1−+−···+(−1)23ny²µ (1)n>m§K|xn−xm|=am+1+am+1qm+1+···+anqn6M|q|m+1+|q|m+2+···+|q|n=m课后答案网+1qn−mm+11−|q|m+11M|q|0,∃N∈Z§n>m>Nž§kM|q|<ε§lk|xn−xm|<ε.1−|q|d…ÜÂñn§{xn}7Âñ.1sin(n+1)sin(n+2)sinm(2)m>n§é∀ε>0£Ø”www.hackshp.cnε<2¤§du|xm−xn|=2n+1+2n+2+···+2m6m−n11−111111121++···+=1++···+=<§e‡|xm−2n+12n+22m2n+122m−n−12n+112n1−21xn|<姕‡<ε=Œ.2nlnε+N=∈Z§m>n>Nž§k|xm−xn|<ε.1ln2d…ÜÂñn§{xn}7Âñ.1£½µ3£1¤¥-a0=1,ak=sink,q=§Kd£1¤=£2¤¤.2(−1)n+2(−1)n+3(−1)n+k−111(−1)k−1(3)é∀ε>0§é∀k∈Z+§du|xn+k−xn|=++···+=−+···+=n+1n+2n+kn+1n+2n+kk111(−1)111−−+···+<<§e‡|xn+k−xn|<姕‡<ε=Œ.n+1n+2n+3n+kn+1nn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn46d…ÜÂñn§limf(xn)•3§Ø”limf(xn)=An→∞n→∞dxn?¿59¼ê4•†ê4•"X•§limf(x)=A=limf(x)•3.x→+∞x→+∞00016.y²limf(x)•3¿©7‡^‡´µé?¿‰½ε>0§•3δ>0§0<|x−x|<δ,0<|x−x|<00x→x0000δž§ðk|f(x)−f(x)|<ε.y²µ⇒®•limf(x)•3§Ø”limf(x)=A.x→xx→x00εé∀ε>0,∃δ>0§0<|x−x|<δž§k|f(x)−A|<020000ε00ε0000<|x−x|<δ,0<|x−x|<δž§k|f(x)−A|<,|f(x)−A|<§K|f(x)−f(x)|=0022000000|f(x)−A−(f(x)−A)|6|f(x)−A|+|f(x)−A|<ε§lé?¿‰½ε>0§•3δ>0§0<000000|x−x|<δ,0<|x−x|<δž§ðk|f(x)−f(x)|<ε.00⇐3f(x)½Â•S§?¿Àê{xn}§¦xn→x…xn6=x(n→∞)000000000d®•§é∀ε>0,∃δ>0§x,x∈D(f)§…0<|x−x|<δ,0<|x−x|<δž§Òk|f(x)−0000f(x)|<ε.+qÏxn→x,xn6=x(n→∞)§u´éþãδ>0§½∃N∈Z§n>Nž§k0<|xn−x|<δ§l000n,m>Nž§Òk0<|xn−x|<δ,0<|xm−x|<δ§?k|f(xn)−f(xm)|<ε.00dê…ÜÂñn§limf(xn)•3§Ø”limf(xn)=An→∞n→∞d{xn}´?¿±x•4•ê…xn6=x9¼ê4•†ê4•"X•§limf(x)=A=limf(x)•00x→xx→x003.00017.y²f(x)3x:ëY¿©7‡^‡´µé?¿‰½ε>0§•3δ>0§|x−x|<δ,|x−x|<δž§ð000000k|f(x)−f(x)|<ε.εy²µ⇒®•f(x)3x:ëY§Ké∀ε>0,∃δ>0§|x−x|<δž§k|f(x)−f(x)|<00020000ε00ε000|x−x|<δ,|x−x|<δž§k|f(x)−f(x)|<,|f(x)−f(x)|<§K|f(x)−f(x)|=000202000000|f(x)−f(x)−(f(x)−f(x))|6|f(x)−f(x)|+|f(x)−f(x)|<ε§lé?¿‰½ε>0§•00000000003δ>0§|x−x|<δ,|x−x|<δž§ðk|f(x)−f(x)|<ε.00000⇐x=x,x=x§Kd®•§é∀ε>0,∃δ>0§|x−x|<δž§Òk|f(x)−f(x)|<ε.000lf(x)3x:ëY.0课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn51)µ√√√√√√∆yx+∆x−x2.01−21(1)===1002.01−2=√√∆x∆x0.012.01+211−∆yx+∆xx1125(2)==−=−=−∆x∆xx(x+∆x)4(4+0.04)4047.y²µ(1)∆(f(x)±g(x))=∆f(x)±∆g(x)(2)∆[f(x)·g(x)]=g(x+∆x)·∆f(x)+f(x)·∆g(x)y²µ(1)∆(f(x)±g(x))=[f(x+∆x)±g(x+∆)]−[f(x)±g(x)]=[f(x+∆x)−f(x)]±[g(x+∆x)−g(x)]=∆f(x)±∆g(x)(2)∆[f(x)·g(x)]=f(x+∆x)·g(x+∆x)−f(x)·g(x)=f(x+∆x)·g(x+∆x)−f(x)·g(x+∆x)+f(x)·g(x+∆x)−f(x)·g(x)=[f(x+∆x)−f(x)]·g(x+∆x)+f(x)[g(x+∆x)−g(x)]=g(x+∆x)·∆f(x)+f(x)·∆g(x)课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn52§2.{ü¼êê1.dê½Â¦y=cosxê.−2sin2x+∆xsin∆xsin∆x0cos(x+∆x)−cosx22∆x2)µy=lim=lim=−limsinx+=∆x→0∆x∆x→0∆x∆x→02∆x20−sinx§=(cosx)=−sinx.√2.dê½Â¦y=3xê."1#"1#1∆x3−2∆x3√√x31+−1x31+−123x+∆x−3xxxx−30)µy=lim=lim=lim==∆x→0∆x∆x→0∆x∆x→0∆x3x1√01√§=(3x)=√33x233x23.U½Ây²µŒó¼êÙ¼ê´Û¼ê§ŒÛ¼êÙ¼ê´ó¼ê.y²µf(x)•Œó¼ê§Kf(−x)=f(x)¶g(x)•ŒÛ¼ê§Kg(−x)=−g(x)0f(−x+∆x)−f(−x)f(x−∆x)−f(x)−[f(x−∆x)−f(x)]u´f(−x)=lim=lim=lim=∆x→0∆x∆x→0∆x−∆x→0−∆x0−f(x)=Œó¼êÙ¼ê´Û¼ê¶0g(−x+∆x)−g(−x)−g(x−∆x)+g(x)g(x−∆x)−g(x)]0g(−x)=lim=lim=lim=g(x)=Œ∆x→0∆x∆x→0∆x−∆x→0−∆xÛ¼êÙ¼ê´ó¼ê.4.U½Ây²µŒ±Ï¼ê§Ù¼êE•±Ù¼ê.y²µf(x)•Œ±Ï•T¼ê§Kf(x+T)=f(x)§0f(x+T+∆x)−f(x+T)f(x+∆x)−f(x)0u´f(x+T)=lim=lim=f(x)=Œ±Ï¼ê§∆x→0∆x∆x→0∆xÙ¼êE•±Ù¼ê.课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn53§3.¦{K1.|^®²‰Ñêúª§¦e¼êêµ5(1)y=x11(2)y=x6(3)y=xx(4)y=2(5)y=log10xx(6)y=10)µ0504(1)y=(x)=5x011010(2)y=(x)=11x0605(3)y=(x)=6x0x0x(4)y=(2)=2ln2001(5)y=(log10x)=xln100x0x(6)y=(10)=10ln102.¦e¼êêµ200(1)f(x)=2x−3x+1§¿¦f(0),f(1)500π(2)f(x)=x+3sinx§¿¦f(0),f2x00(3)f(x)=e+2cosx−7x§¿¦f(0),f(π)2(4)f(x)=4sinx−lnx+xnn−100(5)f(x)=anx+an−1x+···+a1x+a0§¿¦f(0),f(1))µ000(1)f(x)=4x−3§f(0)=−3,f(1)=140400π5π(2)f(x)=5x+3cosx§¿¦f(0)=3,f=课后答案网2160x00π(3)f(x)=e−2sinx−7§¿¦f(0)=−6,f(π)=e−701(4)f(x)=4cosx−+2xxPnn−1n−200(5)f(x)=nanx+(n1)an−1x+···+a1§¿¦f(0)=a1,f(1)=iaii=13.¦e¼êêµwww.hackshp.cn200π(1)y=xsinx§¿¦f(0),f2200(2)y=xcosx+3x§¿¦f(−π)Úf(π)(3)y=xtanx+7x−6x2(4)y=esinx−7cosx+5x√13(5)y=4x+−2xx2(6)y=(3x+2x−1)sinx)µ0200π(1)y=2xsinx+xcosx§f(0)=0,f=π2000(2)y=cosx−xsinx+6x§f(−π)=−1−6π,f(π)=−1+6π02(3)y=tanx+xsecx+70xxx(4)y=esinx+ecosx+7sinx+10x=e(sinx+cosx)+7sinx+10x若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn540212(5)y=√−−6xxx202(6)y=(3x+2x−1)cosx+(6x+2)sinx4.¦e¼êêµ2+sinx(1)y=x(2)y=cotx23x+7x−1(3)y=√x2(1+x)sinx(4)y=2xxlnx(5)y=1+xxxe−1(6)y=sinx)µ00x(2+sinx)−(x+sinx)xcosx−sinx−2(1)y==x2x20cosx0sinx(cosx)0−cosx(sinx)012(2)y===−=−cscxsinxsin2xsin2x√23x+7x−1√20√02x(6x+7)−√20x(3x+7x−1)−(x)(3x+7x−1)2x9x+7x+1(3)y===√=xx2xx917−11−3x2+x2+x222220202x[(1+x)sinx]−2(1+x)sinx(4)y==4x222222x[2xsinx+(1+x)cosx]−2(1+x)sinx(x−1)sinx+x(1+x)cosx=4x22x200(1+x)(xlnx)−xlnx(1+x)(lnx+1)−xlnxx+lnx+1(5)y===(1+x)2(1+x)2(1+x)2x00xxx0sinx(xe−1)−(sinx)(xe−1)esinx(x+1)−cosx(xe−1)(6)y==sin2xsin2x5.¦e¼êêµ课后答案网√x+cosx2(1)y=−7xx−1xsinx+cosx(2)y=xsinx−cosx22x3+x2(3)y=xesinx+√−xlnx+8xxwww.hackshp.cnsinx(4)y=1+tanxxcosx−lnx(5)y=x+11(6)y=x+cosx)µ1√(x−1)(√−sinx)−(x+cosx)02x(1)y=−14x=(x−1)2√√(x−1)(1−2xsinx)−(2x+2xcosx)√−14x2x(x−1)20(xsinx−cosx)(sinx+xcosx−sinx)−(xsinx+cosx)(sinx+xcosx+sinx2(sinxcosx+x)(2)y==−=(xsinx−cosx)2(xsinx−cosx)22x+sin2x−(xsinx−cosx)2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn55√23+x2xx−√0x2x2x2xx(3)y=2xesinx+xesinx+xecosx+−lnx−1+16x=xe(2sinx+xsinx+x23x−1xcosx)+√−lnx−1+16x2xx20cosx(1+tanx)−sinx·secx(4)y=(1+tanx)21(x+1)(cosx−xsinx−)−(xcosx−lnx)0x(5)y==(x+1)22xcosx−(xsinx+1)(x+1)+xlnxx(x+1)201−sinxsinx−1(6)y=−=(x+cosx)2(x+cosx)236.¦•‚y+1=(x−2)3:A(3,0)?ƒ‚•§9{‚•§.33020)µÏy+1=(x−2)§Ky=(x−2)−1§u´y=3(x−2)§K¤¦ƒ‚Ç•k=y|x=3=3§1l¤¦ƒ‚•§•µy=3(x−3)=3x−y−9=0¶¤¦{‚•§•µy=−(x−3)=x+3y−3=0.37.¦•‚y=lnx3:(1,0)?ƒ‚•§Ú{‚•§.010)µÏy=lnx§Ky=§u´¤¦ƒ‚Ç•k=y|x=1=1§xl¤¦ƒ‚•§•µy=x−1=x−y−1=0¶¤¦{‚•§•µy=−(x−1)=x+y−1=0.2o8.Ô‚y=x−2x+43=˜:ƒ‚²1ux¶º3=˜:ƒ‚†x¶•45º20)µÏy=x−2x+4§y=2x−2.q²1ux¶ƒ‚Ç•k=0§K2x−2=0§u´x=1§=¤¦:•(1,3)¶o3313q†x¶•45ƒ‚Ç•k=1§K2x−2=1§u´x=§=¤¦:•,.2244329.÷†‚$ÄÔN§Ù$Ä•§•s=3t−20t+36t§¦Ù„ݧ¿¯ÔNÛž•c$ĺ۞• $ĺ432032)µÏs=3t−20t+36t§v=s=12t−60t+72t.v>0=03ž§ÔN•c$Ķv<0=20=yx+ax+bž§ŠØу‚.00014.¯.êa•ŸoŠž§†‚y=xâU†éê•‚y=logaxƒƒº3Û?ƒƒº00111)µdK¿§x=(logax)§=1=§Kx=§u´y=.xlnalnalna111qdu3ƒ:ƒƒ§Ùp‹I7Lƒ §Klogax=§u´x=e§KŒlna=§=a=ee=.lnae1êa=eež§†‚y=xâU†éê•‚y=logaxƒƒ§3:(e,e)?ƒƒ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn56§4.Eܼê¦{1.¦e¼êêµ(1)y=2sin3x(2)y=4cos(3t−1)2x(3)y=3e+5cos2x2(4)y=(x+1)23(5)y=(1−x+x)−2t(6)y=3e+1(7)y=ln(x+1)4(8)y=(3x+1)√(9)y=1+x221(10)y=1−xx(11)y=tan+sin3x2(12)y=lnsinxx(13)y=√1+x21−3t2(14)y=√e2π)µ0(1)y=6cos3x0(2)y=−12sin(3t−1)02x(3)y=6e−10sin2x0(4)y=2(x+1)022(5)y=3(1−x+x)(2x−1)0−2t(6)y=−6e课后答案网01(7)y=x+103(8)y=12(3x+1)0x(9)y=√1+x20112(x−1)(10)y=21−·−=xx2www.hackshp.cnx3012x(11)y=sec+3cos3x220cosx(12)y==cotxsinx√x21+x2−√01+x21(13)y==231+x(1+x2)2√0−32t−3t2(14)y=√eπ2.¦e¼êêµ3(1)y=sin2x−2t(2)y=(at+b)e(a,b•~ê)22tt(3)y=esin3t+2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn5721−x(4)y=ln1+x2−ktesinωt(5)y=(k,ω•~ê)1+t4(6)y=(x+cos2x)2−t(7)y=e(cost+sint)x(8)y=√1+cos2x√(9)y=(x−1)x2+12(10)y=(2+3t)sin2t+7t−7)µ02(1)y=6sin2xcosx=3sin4xsin2x0−2t−2t−2t(2)y=ae−2(at+b)e=−(2at+2b−a)e02t2t2t(3)y=2esin3t+3ecos3t+t=e(2sin3t+3cos3t)+t22201+x−2x(1+x)−2x(1−x)4x(4)y=·=1−x2(1+x2)2x4−1−kt−kt0(1+t)e(−ksinωt+ωcosωt)−(esinωt(5)y==(1+t)2−kt−kt−(kt+k+1)esinωt+ω(1+t)ecosωt(1+t)22004[(x+cos2x)]8(1−2sin2x)(6)y=−=−(x+cos2x)4(x+cos2x)20−t−t−t(7)y=−e(cost+sint)+e(−sint+cost)=−2esint√−2sinxcosx1+cos2x−x√21+cos2x1+cos2x+xsinxcosx0(8)y==231+cosx(1+cos2x)2√2x2x2−x+1(9)y0=x2+1+(x−1)√=√2x2+1x2+10(10)y=3sin2t+2(2+3t)cos2课后答案网t+14t3.¦e¼êêµ−kt(1)y=e(3cosωt+4sinωt)(k,ω•~ê)(2)y=xarctanx22−x(3)y=(2x+1)esin3x−tesin3twww.hackshp.cn(4)y=√1+t2t(5)y=(3t+1)e(cos3t−7sin3t)−2t(6)y=tarcsin3t+7elnt+8t√x(7)y=xa2−x2+√(a•~ê)a2−x2)µ0−kt−kt−kt(1)y=−ke(3cosωt+4sinωt)+e(−3ωsinωt+4ωcosωt)=e[(4ω−3k)cosωt−(3ω+4k)sinωt]0x(2)y=arctanx+1+x202−x2222−x22(3)y=4x(2x+1)esin3x−(2x+1)e−xsin3x+3(2x+1)e−xcos3x=e(2x+1)[(−2x+28x−1)sin3x+3(2x+1)cos3x]√te−t(−sin3t+3cos3t)1+t2−e−tsin3t√1+t2e−t[3(1+t2)cos3t−(t2+t+1)sin3t]0(4)y==231+t(1+t2)2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn580tttt(5)y=3e(cos3t−7sin3t)+(3t+1)e(cos3t−7sin3t)+(3t+1)e(−3sin3t−21cos3t)=−e[(60t+17)cos3t+(30t+31)sin3t]−2t03t−2t7e(6)y=arcsin3t+√−14elnt++81−9t2t√x2a2−x2+√√x2a2−x2(a2−2x2)(a2−x2)+a2(7)y0=a2−x2−√+=a2−x2a2−x2223(a−x)24.¦e¼êêµn(1)y=sinxcosnxn(2)y=sinhxcoshnx−x2+2x(3)y=en(4)y=(sinx+cosx)(5)y=arcsin(sinx·cosx)r(x+2)(x+3)(6)y=lnx+12x(7)y=arctan1−x2x(8)y=√a2a2+x2)µ0n−1n(1)y=nsinxcosxcosnx−nsinxsinnx=n−1nsinxcos(n+1)x0n−1n(2)y=nsinhxcoshxcoshnx+nsinhxsinhnx=nnsinhxcosh(n+1)x0−x2+2x(3)y=−2(x−1)e0n−1n−2(4)y=n(sinx+cosx)(cosx−sinx)=n(sinx+cosx)cos2x0cos2x2cos2x(5)y=p=√1−(sinx·cosx)24−sin22xr课后答案网(x+2)(x+3)101111(6)Ïy=ln=[ln(x+2)+ln(x+3)−ln(x+1)]§y=+−=x+122x+2x+3x+12x+2x−1.2(x+1)(x+2)(x+3)22012(1−x)+4x2(7)y=2·(1−x2)2=1+x22x1+1−x2www.hackshp.cn√x2a2+x2−√0a2+x21(8)y==a2(a2+x2)223(a+x)25.|^éê2¦•{§¦e¼êêµr1−x(1)y=x1+xr2xx+1(2)y=1−x1+x+x2(3)y=(x−αα1α2αn1)(x−α2)···(x−αn)√(4)y=(x+1+x2)nmx(5)y=xm)µ若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn59r1−x11101−1(1)Ïy=x§Klny=lnx+ln(1−x)−ln(1+x)§ü>éx¦§y=+−1+x22yx2(1−x)1§2(1+x)201−x−xKy=√(0<|x|<1)(1+x)1−x2r2xx+1112(2)Ïy=§Klny=2lnx−ln(1−x)+ln(x+1)−ln(1+x+x)§ü>éx¦1−x1+x+x222102111+2x§y=+++−§yx1−x2(1+x)2(1+x+x2)r20xx+122x+1Ky=+11−x+12(x+1)−1−x1+x+x2x2(1+x+x2)QnQn(3)Ïy=(x−αα1α2αnαiαi1)(x−α2)···(x−αn)=(x−αi)9y3éêÎÒS§Alim(x−αi)>i=1i=1nPn10Xαi0§Klny=αiln|x−αi|§ü>éx¦ê§y=§i=1yx−αii=1n0PnαiYαQnαKy=(x−αi)i(x∈D)Ù¥D=(x−αi)i>0i=1x−αii=1i=1x1+√√n√101+x2(4)Ïy=(x+1+x2)§Klny=nln(x+1+x2)§ü>éx¦§y=n√=pyx+1+x2√n§Ky0=√n(x+1+x2)n1+x21+x2mx10m0m−1x+1(5)Ïy=xm§Klny=mln|x|+xlnm§ü>éx¦§y=+lnm§Ky=xm+yxmxxmlnmdy6.f(x)´éxŒ¦¼ê§¦.dx2(1)y=f(x)xf(x)(2)y=f(e)·e(3)y=f(f(f(x))))µdy02课后答案网(1)=2xf(x)dxdyx0xf(x)0xf(x)f(x)x0xx0(2)=ef(e)·e+f(x)f(e)e=e(ef(e)+f(e)f(x))dxdy000(3)=f(f(f(x)))f(f(x))f(x)dxwww.hackshp.cndy7.ϕ(x),ψ(x)•éxŒ¦¼ê§¦.dxp(1)y=ϕ2(x)+ψ2(x)ϕ(x)(2)y=arctan(ψ(x)6=0)ψ(x)p(3)y=ϕ(x)ψ(x)(ϕ(x)6=0,ψ(x)>0)(4)y=logϕ(x)ψ(x)(ϕ(x)>0,ψ(x)6=0))µ00dyϕ(x)ϕ(x)+ψ(x)ψ(x)(1)=pdxϕ2(x)+ψ2(x)00dyϕ(x)ψ(x)−ψ(x)ϕ(x)(2)=dxϕ2(x)+ψ2(x)dypψ0(x)ϕ0(x)lnψ(x)(3)=ϕ(x)ψ(x)−dxϕ(x)ψ(x)ϕ2(x)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn6000ψ(x)ϕ(x)lnϕ(x)−lnψ(x)00dyψ(x)ϕ(x)ψ(x)ϕ(x)lnψ(x)(4)==−=dx(lnϕ(x))2ψ(x)lnϕ(x)ϕ(x)(lnϕ(x))200ψ(x)ϕ(x)logϕ(x)ψ(x)−ψ(x)lnψ(x)ϕ(x)lnϕ(x)8.¦ã4-7¤«•YëÅw¬$Ä„Ý.pr2ωsin2ωt)µÏs=l2−r2sin2ωt−rcosωt§v=s0=rωsinωt−p.2l2−r2sin2ωt√19.¦•‚y=1−x23x=?ƒ‚•§Ú{‚•§.2√0x13)µÏy=−√§K3x=?ƒ‚Ç•k=−§1−x2√2√3331√u´¤¦ƒ‚•§•µy−=−x−=x+3y−2=0¶232√3√1√¤¦{‚•§•µy−=3x−=3x−y=0.22−x10.¦•‚y=eþ˜:§¦LT:ƒ‚††‚y=−ex²1§¿ÑT:{‚•§.0−x)µÏk=y=−e=−e§Kx=−1§KL(−1,e):ƒ‚††‚y=−ex²1§LT:{‚•§12•y−e=(x+1)=x−ey+e+1=0.e√11.¦•‚y=1−x2þY²ƒ‚.0x)µÏk=y=−√=0§Kx=0§u´d•‚3(0,1)?ƒ‚•Y²ƒ‚§ƒ‚•§•y=1.1−x21p12.¦•‚y=(1+2x2±1+4x2)þî‹Ix=U:?ƒ‚•§.ùƒ‚„†•‚uÛ?º202x2U)µÏy=2x±√§K•‚3x=U?ƒ‚Ç•k=2U±√§u´d•‚3ƒ1+4x21+4U212p212p22U:(U,(1+2U±1+4U))?ƒ‚•§•y−(1+2U±1+4U))=(2U±√)(x−U)§221+4U2√√112p=2U(1+4U2±1)x−1+4U2y±+(1−2U)1+4U2=0§dƒ‚„†•‚u22s√√√U(1+4U2±1)12U2(1+4U2±1)24U2(1+4U2±1)2√,1+±1+.1+4U221+4U21+4U2013.y=f(x)3xŒ§Pϕ(t)=f(x+at)§a•~ꧦϕ(0).000)µea=0§Kϕ(t)=f(x)§Kϕ(0)=00课后答案网0ϕ(x)−ϕ(0)f(x0+at)−f(x0)f(x0+at)−f(x0)0ea6=0§Kϕ(x)=lim=lim=alim=af(x).0t→0tt→0tt→0atwww.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn61§5.‡©9Ù$Ž1.¦e¼ê3•½:‡©µnn−1(1)y=anx+an−1x+···+a0§¦dy(0),dy(1)π(2)y=secx+tanx§¦dy(0),dy,dy(π)41x(3)y=arctan§¦dy(0),dy(a)aa11(4)y=+§¦dy(0.1),dy(0.01)xx2)µPnn−1n−2(1)Ïdy=[nanx+(n−1)an−1x+···+a1]dx§Kdy(0)=a1dx,dy(1)=iaidxi=1π√2(2)Ïdy=(tanxsecx+secx)dx§Kdy(0)=dx,dy=(2+2)dx,dy(π)=dx4dxdxdx(3)Ïdy=dx§Kdy(0)=dx,dy(a)=dxa2+x2a22a2x+2(4)Ïy=−dx§Kdy(0.1)=−2100dx,dy(0.01)=−2010000dxx32.¦e¼êy=y(x)‡©µ121314(1)y=x−x+x−x2342(2)y=xsinxx(3)y=1−x2(4)y=xlnx−x2n(5)y=(1−x)√1(6)y=x+lnx−√x(7)y=lntanx(8)y=sinaxcosbx(9)y=eaxcosbx课后答案网√(10)y=arcsin1−x2)µ23(1)dy=(1−x+x−x)dx2(2)dy=(2xsinx+xcosx)dxwww.hackshp.cn21+x(3)dy=dx(1−x2)2(4)dy=lnxdx2n−1(5)dy=−2nx(1−x)dx√x+2x+1(6)dy=dx3x22(7)dy=dxsin2x(8)dy=(acosaxcosbx−bsinaxsinbx)dxax(9)dy=e(acosbx−bsinbx)dxx(10)dy=−√dx|x|1−x23.¦e¼êy‡©µ2(1)y=sint,t=ln(3x+1)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn622(2)y=ln(3t+1),t=sinx3u12(3)y=e,u=lnt,t=x−2x+5222(4)y=arctanu,u=(lnt),t=1+x−cotx)µ3sin(2ln(3x+1))(1)dy=dx3x+13sin2x(2)y=dx3sin2x+123(3x−2)3ln(x2−2x+5)(3)y=e2dx2(x3−2x+5)222ln(1+x−cotx)(2x+cscx)(4)y=dx[1+(ln(1+x2−cotx))4](1+x2−cotx)4.eu,v,w•xŒ‡©¼ê§¦¼êy‡©dyµ(1)y=u·v·wu·w(2)y=v21(3)y=√u2+v2√(4)y=lnu2+v2u(5)y=arctanv)µ000(1)dy=(u·v·w+u·v·w+u·v·w)dx2000v(uw+uw)−2uvvw(2)dy=dxv400uu+vv22(3)dy=−dx(u+v>0)3(u2+v2)200uu+vv(4)dy=dxu2+v200uv−uv课后答案网(5)dy=dx(v6=0)u2+v2www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn63§6.Û¼ê9ëꕧ¤L«¼ê¦{dy1.¦eÛ¼êêµdx22xy(1)+=1§Ù¥a,b•~êa2b22(2)y=2px§Ù¥p•~ê222(3)x+xy+y=a§Ù¥a•~ê33(4)x+y−xy=01(5)y=x+siny2222(6)x3+y3=a3§Ù¥a•~ê(7)y−cos(x+y)=0(8)y=x+arctanyy(9)y=1−ln(x+y)+eyp(10)arctan=lnx2+y2x)µ022x2yy0bx(1)3•§üàéx¦ê§¿5¿y´x¼ê§Òk+=0§Ky=−(y6=0).a2b2a2y00p(2)3•§üàéx¦ê§¿5¿y´x¼ê§Òk2yy=2p§Ky=(y6=0).y0002x+y(3)3•§üàéx¦ê§¿5¿y´x¼ê§Òk2x+xy+y+2yy=0§Ky=−.x+2y2220003x−y(4)3•§üàéx¦ê§¿5¿y´x¼ê§Òk3x+3yy−xy−y=0§Ky=.x−3y200y02(5)3•§üàéx¦ê§¿5¿y´x¼ê§Òky=1+cosy§Ky=.22−cosyr2−12−100x(6)3•§üàéx¦ê§¿5¿y´x¼ê§Òkx3+y3y=0§Ky=−3.33y(7)3•§üàéx¦ê§¿5课后答案网¿y´x¼ê§Òky0+(1+y0)sin(x+y)=0§Ky0=−sin(x+y).1+sin(x+y)020y01+y(8)3•§üàéx¦ê§¿5¿y´x¼ê§Òky=1+§Ky=.1+y2y2001+y0y01(9)3•§üàéx¦ê§¿5¿y´x¼ê§Òky=−+ye§Ky=.x+y(x+y)ey−x−y−1www.hackshp.cn00xy−yx+yy0x+y(10)3•§üàéx¦ê§¿5¿y´x¼ê§Òk=§Ky=.x2+y2x2+y2x−ydy2.¦eÛ¼ê3•½:êµdx1π(1)y=cosx+siny§:,022x(2)ye+lny=1§:(0,1))µ00y02sinx(1)3•§üàéx¦ê§¿5¿y´x¼ê§Òky=−sinx+cosy§Ky=§u´32cosy−2π0:,0?§y=−2.202xx0y0ye(2)3•§üàéx¦ê§¿5¿y´x¼ê§Òke(y+y)+=0§Ky=−§u´3yyex+101:(0,1)?§y=−.2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn64333.¦•‚x2+y2=163:(4,4)ƒ‚•§Ú{‚•§.r313100x0)µ3•§üàéx¦ê§¿5¿y´x¼ê§Òkx2+y2y=0§Ky=−§u´y|x=4=22yy=4−1§lƒ‚•§•y−4=−(x−4)§=x+y−8=0{‚•§•y−4=x−4§=x=y.4.¦eëꕧ3¤«:êµx=acostππ(1)3t=Ú?y=bsint34x=t−sintπ(2)§3t=,π?y=1−cost2√√223x=1−t(3)3§3t=,?y=t−t23x=a(t−sint)π(4)(a´~ê)§3t=0,?y=a(1−cost)2)µ√000dyy(t)bπ03b(1)Ïx(t)=−asint,y(t)=bcost§K==−cott§u´§t=ž§y=−¶dxx0(t)a33aπ0bt=ž§y=−4a000dyy(t)sintπ0(2)Ïx(t)=1−cost,y(t)=sint§K==§u´§t=ž§y=1¶t=dxx0(t)1−cost20πž§y=0√√02002dyy(t)3t−1202(3)Ïx(t)=−2t,y(t)=1−3t§K==§u´§t=ž§y=¶t=dxx0(t)2t24√30ž§y=03000dyy(t)t0(4)Ïx(t)=a(1−cost),y(t)=asint§K==cot§u´§t=0ž§yÿ¶t=dxx0(t)2πž§y=125.¦eëꕧêµx=acosht(1)课后答案网y=bsinht2x=sint(2)2y=cost3x=acost(3)3y=asint2t2x=ecostwww.hackshp.cn(4)2t2y=esint)µ0dyy(t)asinhta(1)===cothtdxx0(t)bcoshtb0dyy(t)−2costsint(2)===−1dxx0(t)2sintcost02dyy(t)3sintcost(3)===−!tantdxx0(t)−3cos2tsint02t2dyy(t)e(2sint+2sintcost)sint+cost(4)===tant·dxx0(t)e2t(2cos2t−2costsint)cost−sint6.˜I/Nì§ 10º§þºŒ»•4º£ã4-11¤µ(1)/Yž§¦YNÈVéY¡pÝhCzǶ(2)¦NÈVéNì ¡Œ»RCzÇ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn6512Rh5)µÏNÈV†Nì ¡Œ»R§Y¡pÝh"X•V=πRh§…d®•§==h=R§u34102´21243dV42(1)V=πhh=πh§l=πh¶3575dh2512553dV52(2)V=πR·R=πR§l=πR.326dR237.˜I/Nì.¡Šþ˜X§§º•2arctan§8•p¡?,«—N§4dr1dV(1)—NŒ»r•3§Œ»O„Ý•ž§NÈO„Ý´õºdt4dt(2)—NŒ»•6§NÈO„Ý•24ž§Œ»O„Ý´õº43dV42drdV)µÏNÈV†—NŒ»r"X•V=πr§V,rÑ´žmt¼ê§ü>ét¦§=π(3r)==9dt9dtdt42drπr§K3dtdr1dV(1)r=3,=ž§=3π¶dt4dtdr3dVdVdr1(2)d=§r=6,=24ž§=.dt4πr2dtdtdt2π8.Ylp•18f’!.Œ»•6f’I/¦Ì6Œ»•5f’Î/ÙS.®•¦Ì¥Y •12f’ž§¦Ì¥Y¡eü„Ý•1f’/©§¦džÙ¥Y¡þ,„Ý.)µlm©¦Yå²t©¨ §I/¦Ì¥M— Ý•xf’§Î/Ù¥Y¡,pyf’"dž§121x2π3¦Ì¥¦ÑM—NÈ•π·6·18−π·6·x=216π−x£á•f’¤§Î/Ù¥533182732π31xM—NÈ•π·5·y=25πy£á•f’¤"âK¿•§25πy=216π−x§y=216−§u272527dydydx12dx12dxdx´=·=−·3x·=−x·"x=12(f’)ž§=−1(f’/©)§u´džÙdtdxdt675dt225dtdtdy1216¥Y¡þ,„Ý•=−·12(−1)==0.64(f’/©).dx225252UdP9.ã4-12¤«>´¥§ÑÑõÇP=iR§Ù¥>6i=.¦NŒC>{Rž§õÇPCzÇ.r+RdR2222UdPdi2−2URUU(r−R))µÏP=iR,i=§K=2iR+i=+=r+RdR课后答案网dR(r+R)3(r+R)2(r+R)3www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn66§7.ØŒ¼êÞ~001.¦e¼ê3¤«:x0†êf−(x0)Úmêf+(x0)µ2x,x60,(1)y=xx0=0xe,x>0,(x,x6=0,1(2)y=1+exx0=00,x=0,(21xsin,x6=0,(3)y=xx0=00,x=0,)µx20xe−00x−0(1)f+(x0)=lim=1¶f−(x0)=lim=0.x→+0xx→−0xx−0101+ex1(2)f+(x0)=lim=lim1=0¶x→+0xx→+01+ex01f−(x0)=lim1=1.x→−01+ex2121xsin−0xsin−0(3)f0(x)=limx=0¶f0(x)=limx=0.+0−0x→+0xx→−0x2.¦e¼ê3êØ•3:†!mêµ(1)y=|ln|x||(2)y=|tanx|√(3)y=1−cosx)µln(−x),x6−1−ln(−x),−11ddŒ•§¼ê3x=0,x=±1?êØ•3"0−ln[−(−1+∆x)]−ln(−(−1))−1f+(−1)=lim=limln(1−∆x)∆x=1¶∆x→+0∆x∆x→+00−ln[−(−1+∆x)]−ln(−(−1))−1f−(−1)=lim=limln(1−∆x)∆x=−1¶∆x→−0∆x∆x→−000ϼê3x=0:ÿ§f+(0)Úf−(0)ÿ¶0ln(1+∆x)−www.hackshp.cnln11f+(1)=lim=limln(1+∆x)∆x=1¶∆x→+0∆x∆x→+00−ln(1+∆x)−ln(1)1f−(1)=lim=lim−ln(1+∆x)∆x=−1.∆x→−0∆x∆x→−0iπ−tanx,x∈kπ−,kπ(2)y=|tanx|=2πk∈Ztanx,x∈kπ,kπ+2πÙ¥x=kπ+(k∈Z)ž¼êý§…•Ã¡mä:§†!mêÿ¶2x=kπ(k∈Z)•êØ•3:.0tan(kπ+∆x)−(−tankπ)tan∆x0−tan(kπ+∆x)−(−tankπ)f+(kπ)=lim=lim=1¶f−(kπ)=lim=∆x→+0∆x∆x→+0∆x∆x→−0∆xtan∆xlim−=−1.∆x→−0∆x0sinx√(3)Ïy=√x6=2kπ(k∈Z)žâk½Â§x=2kπ(k∈Z)•y=1−cosxØŒ:.1−cosxp√√r01+cos(2kπ+∆x)−1+2kπ1−cos∆x1−cos∆xf+(2kπ)=lim=lim=−lim=∆x→+0∆x∆x→+0∆x∆x→+0∆x2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn67√2¶2p√√r01+cos(2kπ+∆x)−1+2kπ1−cos∆x1−cos∆xf−(2kπ)=lim=lim=−lim−=√∆x→−0∆x∆x→−0∆x∆x→−0∆x22−.23.e(1)f(x)3x:Œ§g(x)3x:ØŒ§y²¼êF(x)=f(x)+g(x)3x:ØŒ¶000(2)f(x)Úg(x)3x:ÑØŒ§UÄ佦‚Ú¼êF(x)=f(x)+g(x)3x:ØŒº00y²µ(1)bF(x)=f(x)+g(x)3x:Œ§qf(x)3x:Œ§Kg(x)=F(x)−f(x)3x:Œ§ù†®000•gñ§bؤá"l¼êF(x)=f(x)+g(x)3x:ØŒ.0(2)ØU"~µ|x|+xx−|x|(i)Œµf(x)=,g(x)=3x=0:ÑØŒ§§‚Ú¼êF(x)=f(x)+g(x)=220x3x=0:Œ…F(0)=1¶|x||x|(ii)ØŒµf(x)=,g(x)=3x=0:ÑØŒ§§‚Ú¼êF(x)=f(x)+g(x)=|x|3x=220:•ØŒ.4.3þK^‡e§§‚ÈG(x)=f(x)·g(x)Œœ¹Nº)µ(1)§‚ÈG(x)=f(x)·g(x)3x:ŒUŒ"0~µ0(i)Œµf(x)=x3x=0:Œ…f(0)=1¶g(x)=|x|3x=0:ØŒ§§‚ÈG(x)=0∆G(x)∆x|∆x|f(x)·g(x)=x|x|3x=0Œ…G(0)=lim=lim=lim|∆x|=0∆x→0∆x∆x→0∆x∆x→00(ii)ØŒµf(x)=13x=0:Œ…f(0)=0¶g(x)=|x|3x=0:ØŒ§§‚ÈG(x)=f(x)·g(x)=|x|3x=0:ØŒ.(2)§‚ÈG(x)=f(x)·g(x)3x:ŒUŒ"0~µ2(i)Œµf(x)=|x|,g(x)=|x|3x=0:ÑØŒ§§‚ÈG(x)=f(x)·g(x)=x3x=0Œ0…G(0)=021(ii)ØŒµf(x)=x3,g(课后答案网x)=|x3|3x=0:ÑØŒ§§‚ÈG(x)=f(x)·g(x)=|x|3x=0:ØŒ.015.e¼êf(x)3k•«m(a,b)¥k꧅limf(x)=∞§´Ä7klimf(x)=∞º±~ff(x)=+x→ax→ax1cos`²ƒ.x√‡ƒ§ef(x)3k•«m(a,b)¥k꧅limf0(x)=∞§´Ä7klimf(x)=∞º±~ff(x)=3x`www.hackshp.cnx→ax→a²ƒ.)µ0(1)˜„/`§ØUyklimf(x)=∞.x→aπ11~µéu0,S½Â¼êf(x)=+cos§w,klimf(x)=∞.2xxx→0011111qf(x)=−−−sin=sin−1§x2x2xx2x100éuAϘGêxn=π(n=1,2,···)§kf(xn)=0§limf(xn)=0¶2nπ+n→∞2010022000éuxn=(n=1,2,···)§kf(xn)=−nπ§limf(xn)=−∞§f(x)3x=0:4•Ø•nπn→∞03§•šÃ¡§=limf(x)=∞ؤá.x→0(2)ØUy7klimf(x)=∞.x→a√30√10~µf(x)=x§§3(0,b)(b>0)þk꧅f(x)=§limf(x)=∞§limf(x)=0.33x2x→0x→0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn686.e(1)f(x)3x=g(x)kê§g(x)3x:vkê¶00(2)f(x)3x=g(x)vkê§g(x)3x:kê¶00(3)f(x)3x=g(x)vkê§g(x)3x:•vkê¶00KEܼêF(x)=f(g(x))3x:´ÄŒº0)µ(1)EܼêF(x)=f(g(x))3x:ŒUŒ.0~µ220(i)Œµf(u)=u,g(x)=|x|,x=0§f(u)=u3u=0=g(x)Œ…f(0)=0§g(x)=000220|x|3x=0ØŒ¶F(x)=f(g(x))=|x|=x3x=0Œ…F(0)=0¶000(ii)Œµf(u)=u,g(x)=|x|,x=0§f(u)=u3u=0=g(x)Œ…f(0)=1§g(x)=000|x|3x=0ØŒ¶F(x)=f(g(x))=|x|3x=0ØŒ.00(2)EܼêF(x)=f(g(x))3x:ŒUŒ.0~µ22(i)Œµf(u)=|u|,g(x)=x,x=0§f(u)=|u|3u=0=g(x)ØŒ§g(x)=x3x=0Œ00000220…g(0)=0¶F(x)=f(g(x))=|x|=x3x=0Œ…F(0)=0¶0(ii)Œµf(u)=|u|,g(x)=x,x=0§f(u)=|u|3u=0=g(x)ØŒ§g(x)=x3x=0Œ00000…g(0)=1¶F(x)=f(g(x))=|x|3x=0ØŒ.0(3)EܼêF(x)=f(g(x))3x:ŒUŒ.0~µ2|x|(i)Œµf(u)=2u+|u|,g(x)=x−,x=0§f(u)=2u+|u|3u=0=g(x)ØŒ§g(x)=330002|x|2|x|2|x|x,x>0x−3x=0ØŒ¶F(x)=f(g(x))=2x−+x−==3303333x,x<00x=F(x)=x(∀x∈(−∞,+∞)§F(x)3x=0Œ…F(0)=1¶0(ii)Œµf(u)=|u|,g(x)=|x|,x=0§f(u)=|u|3u=0=g(x)ØŒ§g(x)=|x|3x=0Ø0000Œ¶F(x)=f(g(x))=|x|3x=0ØŒ.0课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn69§8.p ê†p ‡©32000(3)(4)1.y=2x+x+x+1§¦y,y,yÚy.0200(3)(4))µy=6x+2x+1,y=12x+2,y=12,y=0αt00(3)(n)2.y=e(α•~ê)§¦y,y,y.0αt002αt(3)3αt(n)nαt)µy=αe,y=αe,y=αe,y=αe3.¦e¼êp êµx00(1)y=√§¦y1−x200(2)y=xlnx§¦y−x200(3)y=e§¦yarcsinx00(4)y=√§¦y1−x222x000(5)y=x·e§¦y3x000(6)y=a§¦y3(30)(7)y=xsinhx§¦y3(50)(8)y=xcosx§¦y)µ√x21−x2+√01−x212−3002−5(1)y===(1−x)2,y=3x(1−x)2231−x(1−x2)20001(2)y=1+lnx,y=x0−x200−x22−x22(3)y=−2xe,y=−2e(1−2x)=2e(2x−1)xarcsinx1+√01−x21xarcsinx(4)y==+,1−x21−x223(1−x)2x2321arcsinx+√(1−x)2+3x(1−x)2·xarcsinx2x1−x23x(2x2+1)arcsinx00y=+=+(1−x2)2(1−x2)3(1−x2)225课后答案网(1−x)200022x00022x000202x002002x020002x2x2(5)y=(xe)=x(e)+3(x)(e)+3(x)(e)+(x)e=4e(2x+6x+3)03x0023x00033x(6)y=3alna,y=9lna·a,y=27lna·a(7)Ï(x3)0=3x2,(x3)00=6x,(x3)000=6,(x3)(4)=···=(x3)(30)=0;(sinhx)(30)=sinhx,(sinhx)(29)=(28)(27)(30)3(30)3(30)30(29)coshx,(sinhx)=sinhx,(sinhx)=coshx§y=(xsinhx)=x(sinhx)+30(x)(sinhx)+300(28)3000(27)22435(x)(sinhx)+4060(x)(sinh)=xsinhx(x+2610)+30coshx(3x+812)30230030003(4)3(50)(50)(49)(8)Ï(x)=3x,(x)=6x,(xwww.hackshp.cn)=6,(x)=···=(x)=0;(cosx)=−cosx,(cosx)=(48)(47)(50)3(50)3(50)30(49)−sinx,(cosx)=cosx,(cosx)=sinx§y=(xcosx)=x(cosx)+50(x)(cosx)+300(48)3000(47)221225(x)(cosx)+19600(x)(cosx)=xcosx(7350−x)+150sinx(784−x)4.|^êÆ8B{y²e¡úªµx(n)xn(1)(a)=a·(lna)(a>0)(n)π(2)(cosx)=cosx+n·2n−1(n)(−1)·(n−1)!(3)(lnx)=xny²µx0xx1(1)(i)n=1ž§(a)=alna=a(lna)§Kn=1žúª¤á.x(k)xk(ii)bn=kžúª¤á§=(a)=a(lna)¤á§hi0x(k+1)x(k)kx0kxxk+1Kn=k+1ž§(a)=(a)(lna)=(lna)(a)=(lna)·alna=a(lna)§u´n=k+1žúª•¤á.x(n)xnnÜþ㌕§n•?¿g,ꞧúª(a)=a·(lna)(a>0)Ѥá.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn700π(2)(i)n=1ž§(cosx)=−sinx=cosx+§Kn=1žúª¤á.2(k)π(ii)bn=kžúª¤á§=(cosx)=cosx+k·¤á§hi02hπi0π(k+1)(k)Kn=k+1ž§(cosx)=(cosx)=cosx+k·=−sinx+k·=22πcosx+(k+1)·§u´n=k+1žúª•¤á.2(n)πnÜþ㌕§n•?¿g,ꞧúª(cosx)=cosx+n·Ñ¤á.21−101(−1)(1−1)!(3)(i)n=1ž§(lnx)==§Kn=1žúª¤á.xx1k−1(k)(−1)·(k−1)!(ii)bn=kžúª¤á§=(lnx)=¤á§xnhi0(−1)k−1·(k−1)!0(−1)k−1·(k−1)!(k+1)(k)Kn=k+1ž§(lnx)=(lnx)==−k·=xkxk+1kk+1−1(−1)·k!(−1)·(k+1−1)!=§u´n=k+1žúª•¤á.xk+1xk+1n−1(n)(−1)·(n−1)!nÜþ㌕§n•?¿g,ꞧúª(lnx)=Ѥá.xn5.¦n êµ1(1)y=x(1−x)1(2)y=x2−2x−8mm−1(3)y=amx+am−1x+···+a02(4)y=cosωxxe(5)y=xx(6)y=2·lnxax(7)y=epn(x)§Ù¥pn(x)•ngõ‘ª.)µ(n)(n)(n)111课后答案网(n)1111−1(n)(1)Ïy==+§Ky=−=+=(x)+[(1−x(1−x)x1−xx1−xx1−x−1(n)n−n−12n−n−1n−n−1−n−1x)]=(−1)·n!·x+(−1)·n!·(1−x)=n11111(n)111(2)Ïy===−§Ky=−=x2−2x−8(x+2)(x−4)6x−4x+26x−4x+2n1−1(n)−1(n)1n−n−1n−n−1(−1)11[((x−4))−((x+2))]=[(−1)·n!(x−4)−(−1)·n!(n+2)]=n!−6www.hackshp.cn66(x−4)n+1(x+2)n+1(n)2(n)m(n)(n)(3)n>mž§x=(x)=···=(x)=0§Ky=0¶(n)2(n)m−1(n)m(n)m(m)(n)n=mž§x=(x)=···=(x)=0,(x)=(x)=m!§Ky=am·m!¶(n)2(n)n−1(n)n(n)m(n)m!m−nn0)§¦dy3(5)y=lnu(x)§¦dy3(6)y=sin(u(x))§¦dy)µ00(1)dy=(u(x)v(x)+u(x)v(x))dx,20000002dy=[u(x)v(x)+2u(x)v(x)+u(x)v(x)]dx00u(x)v(x)−u(x)v(x)(2)dy=dx,v2(x)000000022u(x)u(x)v(x)+2u(x)v(x)2u(x)(v(x))2dy=−+dxv(x)v2(x)v3(x)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn75m−1n0mn−10(3)dy=[mu(x)v(x)u(x)+nu(x)v(x)v(x)]dx,2m−2n02m−1n−100m−1n00dy=[m(m−1)u(x)v(x)(u(x))+2mnu(x)v(x)u(x)v(x)+mu(x)v(x)u(x)+mn−202mn−1002n(n−1)u(x)v(x)(v(x))+nu(x)v(x)v(x)]dxu(x)0(4)dy=alna·u(x)dx,2u(x)02002dy=alna[lna(u(x))+u(x)]dx00002u(x)2u(x)(u(x))2(5)dy=dx,dy=−dx,u(x)u(x)u2(x)000000033u(x)3u(x)u(x)2(u(x))3dy=−+dxu(x)u2(x)u3(x)0200022(6)dy=cos(u(x))u(x)dx,dy=[cos(u(x))u(x)−sin(u(x))(u(x))]dx,3000000033dy=[cos(u(x))u(x)−3sin(u(x))u(x)u(x)−cos(u(x))(u(x))]dx.课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn761ÊÙ‡©ÆÄ ½n9ÙA^§1.¥Š½n1.3¤ê½n¥§ex•«mà:§ÁÞ~`²(Øؤá.0)µ~µ¼êy=x3«m[−1,1]þk½Â§…Œ§3à:x=1ˆ•ŒŠ§=∀x∈[−1,1]§ðkf(x)600f(x)=1§,y|=16=0.0x=102.éux∈(a,b)§ef(x)>0§K•3§†!m•O−(x,δ),O+(x,δ)¦x∈O−(x,δ)žÿf(x)>000000f(x)§x∈O+(x,δ)žÿf(x)0§â4•5Ÿ§•3xδ(δ>0)•O(x,δ)⊂(a,b)§000x→x0x−x0f(x)−f(x)¦x∈O(x,δ)ž§k0>0§lx∈O−(x,δ)=x−x<0ž§kf(x)>f(x)§0x−x0000x∈O+(x,δ)=x−x>0ž§kf(x)0,f−(x0)<0§K•3x0˜‡•§¦3d•Sf(x)>f(x0).0f(x)−f(x0)y²µÏf+(x0)=lim>0§Kdm4•5Ÿ§7•3x0δ1(δ1>0)m•O+(x0,δ1)§x→x0+0x−x0f(x)−f(x)¦x∈O0+(x,δ1)=00§lkf(x)0)†•O−(x0,δ2)§¦x→x0−0x−x0f(x)−f(x)x∈O0−(x,δ2)=0f(x).00004.ef(x)3[a,b]ëY§f(a)=f(b)=0,f(a)·f(b)>0§Kf(x)3(a,b)S–k˜‡":.000000y²µÏf(a)·f(b)>0§Ø”f(a)>0,f(b)>0(f(a)<0,f(b)<0œ¹ÓnŒy)00f(x)−f(a)qf(a)=f+(a)=lim>0§Kdm4•5Ÿ§7•3aδ1(δ1>0)m•O+(a,δ1)§¦x→a+0x−af(x)−f(a)x∈O+(a,δ1)=00§lkf(a)f(a)qf(a)=0§Kf(x1)>000f(x)−f(b)qf(b)=f−(b)=lim>0§Kd†4•5Ÿ§7•3bδ2(δ2>0)†•O−(b,δ2)§¦x→b−0x−b课后答案网f(x)−f(b)x∈O−(b,δ2)=00§lkf(b)>f(x)¶x−x0½x2∈O−(b,δ1)§Kkf(x2)0,f(x2)<0§Kd":•3½nŒ•§3[x1,x2]S–k˜‡":§q[x1,x2]⊂[a,b]§lf(x)3[a,b]S–k˜‡":.Ón§f0(a)<0,f0(b)<0ž§f(www.hackshp.cnx)3[a,b]S–k˜‡":.05.df(x+∆x)−f(x)=f(x+θ∆x)∆x(0<θ<1)§¦¼êθ=θ(x,∆x)§2(1)f(x)=ax+bx+c(a6=0)1(2)f(x)=xx(3)f(x)=e)µ00(1)f(x)=2ax+b,f(x+θ∆x)=2a(x+θ∆x)+b§22K[2a(x+θ∆x)+b]∆x=f(x+∆x)−f(x)=a(x+∆x)+b(x+∆x)+c−(ax+bx+c)=2112ax·∆x+a(∆x)+b∆x=2ax+∆x+b∆x§u´θ=220101(2)f(x)=−,f(x+θ∆x)=−§x2(x+θ∆x)2∆x11∆x22K−=f(x+∆x)−f(x)=−=−§l∆xθ+2x·∆xθ=0§(x+θ∆x)2x+∆xxx(x+∆x)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn77√−x±x2+x∆x∆x2u´θ=§d?KÒ‡À(θ∈(0,1)½§…Ak>−1(x6=0)(dx+∆xx∆xx∆x>0§K>−1)x0x0x+θ∆x(3)f(x)=e,f(x+θ∆x)=e§x+θ∆xx+∆xxx∆xθ∆x∆xKe∆x=f(x+∆x)−f(x)=e−e=e(e−1)§le∆x=e−1§u´θ=∆x1e−1ln§Œ±yθ∈(0,1)∆x∆x6.f(x)3«m[a,b]SëY§3(a,b)Œ§|^¼êxf(x)1Φ(x)=bf(b)1af(a)1y².‚KFúª§¿Qã¼êΦ(x)AÛ¿Â.xf(x)1y²µÏΦ(x)=bf(b)1=(a−b)f(x)+(f(b)−f(a))x+bf(a)−af(b)§af(a)1qf(x)3«m[a,b]SëY§KdëY¼êoK$Ž{K§•Φ(x)3[a,b]ëY¶qf(x)3(a,b)Œ§KdŒ¼êoK$Ž{K§•Φ(x)3(a,b)Œ.af(a)1bf(b)1qΦ(a)=bf(b)1=0,Φ(b)=bf(b)1=0§Kdâ½n§3(a,b)S–k˜:ξ§af(a)1af(a)10¦Φ(ξ)=0.00000f(b)−f(a)Φ(x)=(a−b)f(x)+f(b)−f(a)§K0=Φ(ξ)=(a−b)f(ξ)+f(b)−f(a)=f(ξ)=.b−ax1y111xΦ(x)Aۿµn/¡ÈúªS∆=2y21§Ù¥(x1,y1),(x2,y2),(x3,y3)L«º:‹I§2x3y31KΦ(x)L«±A(x,f(x)),B(a,f(a)),C(b,f(b))•º:n/¡Èü.07.Áée¼êÑ.‚KFúªf(b)−f(a)=f(c)(b−a)§¿¦c.3(1)f(x)=x,x∈[0,1](2)f(x)=arctanx,x∈[0,1])µ课后答案网√0223323(1)Ïf(x)=3x§K3c(1−0)=1−0=3c=1§qc∈(0,1)§c=.3r0111π4(2)Ïf(x)=§K(1−0)=arctan1−arctan0==§qc∈(0,1)§c=−1.1+x21+c21+c24π0f(b)−f(a)f(c)8.Áée¼êÑ…Üúª=§¿¦c.g(b)−gwww.hackshp.cn(a)g0(c)hiπ(1)f(x)=sinx,g(x)=cosx,x∈0,2√2(2)f(x)=x,g(x)=x,x∈[1,4])µπf−f(0)0002f(c)1−0cosc(1)Ïf(x)=cosx,g(x)=−sinx§K===§½=cotc=1§qc∈gπ−g(0)g0(c)0−1sinc2ππ0,§c=.240001f(4)−f(1)f(c)16−12c3(2)Ïf(x)=2x,g(x)=√§K===§½=4c2=15§qc∈(1,4)§2xg(4)−g(1)g0(c)2−11√2c2153c=.4若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn789.ÁŠ¼êy=|x−1|3«m[0,3]þã/§ùp•Ÿovk²1uuƒ‚§.‚KF½n¥=‡^‡Ø¤áº)µ¼ê3:x=1?ØŒ§=Ùã/ACB•˜ò‚§dò‚3C(0,1):ƒ‚Ø•3§.‚KF½n¥1‡^‡=3(0,3)SŒù˜^‡Ø÷v.y6@@ @ @ @ @ @ @ -03x10.|^.‚KFúªy²Ø ªµ(1)|sinx−siny|6|x−y|ππ(2)x∈−,ž§|x|6|tanx|( Ò•k3x=0ž¤á)22n−1nnn−1(3)n·y(y−x)1,x>y)x(4)0)1+xx(5)ex6=0,e>1+x(©x>0,x<0ü«œ¹y²)y²µ(1)Ø”x>y,f(t)=sint3[y,x]þëY§3(y,x)SŒ§.‚KF½n¤á§Ïksinx−siny=cosξ(x−y)(ξ∈(y,x))§K|sinx−siny|=|cosξ(x−y)|=|cosξ||(x−y)|6|x−y|(∀(x,y∈(−∞,+∞))¤á.π(2)Ø”x∈0,,f(t)=tant3[0,x]þëY§3(0,x)SŒ§.‚KF½n¤á§Ïktanx−22pi2tan0=secξ(x−0)ξ∈(0,x),x∈0,§Kx=cosξ·tanx1§Ky<ξ0)=0)¤á.1+x1+ξ1+xt(5)f(t)=ew,÷v.‚KF½n^‡.tx0ξxξx>0ž§éf(t)=e3[0,x]A^.‚KFúª§ke−e=e(x−0)=e−1=xe(0<ξ1§le−1=xe>x=e>1+x¶t0xξxξx<0ž§éf(t)=e3[x,0]A^.‚KFúª§ke−e=e(0−x)=1−e=−xe(x<ξ<0)§ξxξxÏx<ξ<0§K01+x.xoƒ§ex6=0§oke>1+x.011.ef(x)≡k§Áyf(x)=kx+b.y²µ•ÄF(x)=f(x)−kx00duF(x)=f(x)−k≡0§â.‚KF½níØ1•§F(x)=f(x)−kx=b(∀x∈(−∞,+∞)§f(x)=kx+b.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn79312.y²•§x−3x+c=03[0,1]Sعkü‡ØÓŠ.3y²µ-f(x)=x−3x+c^‡y{.f(x)3[0,1]Skü‡ØÓŠ0|ϕ(x)|,f(x)6=0§K|∆f(x)|>|∆ϕ(x)|.¿y3,xþ∆arctanx6∆ln(1+x)§d212πdy²3,1þ±eØ ª¤áµarctanx−ln(1+x)>−ln2.24000y²µÏ3[a,b]þ|f(x)|>|ϕ(x)|,f(x)6=0§f(x),ϕ(x)3[a,b]þŒ§l3[a,b]þëY.0?x,x+∆x∈[a,b],∆x>0§Kf(x),ϕ(x)3[x,x+∆x]þëYŒ…f(x)6=0.ϕ(x+∆x)−ϕ(x)ϕ0(ξ)∆ϕ(x)ϕ0(ξ)∆ϕ(x)d…ܽn§7•3ξ∈(x,x+∆x)§¦===§u´=f(x+∆x)−f(x)f0(ξ)∆f(x)f0(ξ)∆f(x)ϕ0(ξ)61=|∆f(x)|>|∆ϕ(x)|.f0(ξ)01202x1202xÏ(arctanx)=,(ln(1+x))=§…3,xþ§k2x>1§K(ln(1+x))=>1+x21+x221+x21021=(arctanx)>0§f(x)=ln(1+x),ϕ(x)=arctanx§Kdþ¡(Ø•§3,xþ§∆arctanx=1+x2222|∆arctanx|6|∆ln(1+x)|=∆ln(1+x).1223,1þ?˜‡x§3[x,1]þkarctan1−arctanx=∆arctanx6∆ln(1+x)=ln(1+1)−ln(1+22π22πx)=−arctanx6ln2−ln(1+x)§larctanx−ln(1+x)>−ln2.44014.ef(x)3«mX(d¡½Ã¡)¥äkk.ê§=|f(x)|6M§Kf(x)3X¥˜—ëY.0y²µÏef(x)3«mXþŒ§l•3XþëY§…|f(x)|6M,M>0?x1,x2∈X§Ø”x10§δ=§K|x2−x1|<δ=ž§|f(x2)−f(x1)|6M(x2−x1)0)xy²µ00(1)Ïy=f(x)3(−∞,+∞)SëYŒ§f(x)=1−cosx¶q−16cosx61§f(x)>0§u´y=x−sinx3(−∞,+∞)üNþ,.xx1011(2)Ïy=1+§y=1+ln(1+x)−lnx−xx1+xx1qx>0§1+>0§K•Iä•)Ò¥ªfÎÒ.x11-f(x)=lnx3[x,1+x](é∀x>0)þA^.‚KF½n§kln(1+x)−lnx=(1+x−x)=(x<ξ<ξξ1111111+x)§u´>>§ln(1+x)−lnx=>=ln(1+x)−lnx−>0(∀x>0)§xξ1+xξ1+x1+xx01ddŒ•y>0§ly=1+3(0,+∞)þüNO.x2.üN¼êê´Ä7•üNº)µØ˜½.302~µy=x3(−∞,+∞)þüNþ,§y=3x%ØüN.3.y²eØ ªµ2π(1)x>sinx>x0sinx>x(x<0)62x(3)x−0)23xπ(4)tanx>x+03−(x>1)x课后答案网1pp(6)6x+(1−x)61(06x61,p>1)2p−1y²µππ(1)f(x)=x−sinx§d11K§•f(x)30,SüNþ,§qf(0)=0§é∀x∈0,§kf(x)>22πf(0)=0=x−sinx>0§lx>sinx0g==>,x∈0x§22πxπ2π2πlx>sinx>x00§lx0§x∈(−∞,0)ž§h(x)ü20Nþ,=h(x)g(0)=0=x−−sinx>0(x<0)§Kx−>sinx§lx−>sinx>x(x<0)666若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn852x01x(3)f(x)=ln(1+x)−x,g(x)=ln(1+x)−x+(x>0)§f(x)=−1=−<0(x>0)§21+x1+xKf(x)3(0,+∞)SüNeü§qf(0)=0§é∀x>0§kf(x)0)¶201xg(x)=−1+x=>0(x>0)1+x1+x2xg(x)3(0,+∞)þüNþ,§qg(0)=0§u´g(x)>g(0)=0=ln(1+x)>x−(x>0)§l22xx−0)23xπ02222(4)f(x)=tanx−x−00000f(0)=0=tanx−x−>00x+01)§f(x)=√−=>0(x>1)§u´f(x)3(1,+∞)þüxxx2x2√11ppNþ,§qf(1)=0§u´f(x)>f(1)=0=2x−3+>0(x>1)§l6x+(1−x)6x2p−11(06x61,p>1)pp0p−1p−1(6)f(x)=x+(1−x)(06x61,p>1)§f(x)=px−p(1−x)§0p−1p−1111-f(x)=px−p(1−x)=0§)x=§"f(0)=1,f(1)=1,f=§dd222p−111ppminf(x)=,maxf(x)=1§l6x+(1−x)61(06x61,p>1)06x612p−106x612p−14.(½e¼êþ,!eü«mµ3(1)y=x−6x32(2)y=2x−3x−12x+143(3)y=x−2x(4)y=x+sinx2x(5)y=1+x2课后答案网2(6)y=2x−sinxn−x(7)y=xe(n>0,x60))µ√022(1)Ïy=3x√−6=3(√x−2)§www.hackshp.cn7:x=±2√√00x<−2½x>√2žS§y√>0§¼êî‚þ,¶−22ž§y>0§¼êî‚þ,¶−1ž§y>0§¼êî‚þ,¶x<ž§y60…=3x=0?y=0§¼êî‚eü.2233l3«m,+∞þ¼êî‚þ,¶3«m−∞,þ¼êî‚eü.220(4)Ïy=1+cosx60§¼ê3(−∞,+∞)þ¼êþ,.202(1−x)(5)Ïy=§7:x=±1(1+x2)200x<−1½x>1ž§y<0§¼êî‚eü¶−10§¼êî‚þ,.Sl3«m(−∞,−1)(1,+∞)þ¼êî‚eü¶3«m(−1,1)þ¼êî‚þ,.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn860000(6)Ïy=4x−cosx,y=4+sinx>0§Ky3(−∞,+∞)þüNþ,.00ππ0qy(0)=−1,y=2π§K30,Sk˜‡:x÷vy(x)=0=4x=cosx22000000x>xž§y>0§¼êî‚þ,¶x0,x>0§x>0,e>0§Kxe>00000§¼êî‚þ,¶x>nž§y<0§¼êî‚eü.l3«m(0,n)þ¼êî‚þ,¶3«m(n,+∞)þ¼êî‚eü.5.¦e¼ê4Šµ(1)y=x−ln(1+x)√(2)y=xlnx1(3)y=x+x33(4)y=sinx+cosx(5)y=cosx+coshx)µ01x001(1)Ïy=1−=,y=>01+x1+x(1+x)2d¼ê½Â••(−1,+∞)§K7:•x=0§¼ê•U3ù:k4Š§u´x=0´¼ê4:§4Š•y=0.011100111lnx(2)Ïy=√+√lnx=√(lnx+2),y=−+−lnx=−3333x2x2x2x22x24x222−200−27:•x=e§¼ê•U3ù:k4Š§qy|x=e−2>0§u´x=e´¼ê4:§4Š2•y=−.e01001(3)Ïy=1−,y=−>0x2x3S00d¼ê½Â••(−∞,0)(0,+∞)§K7:•x=±1§¼ê•U3ùü:k4Š§qy|x=1=1>000,y|x=−1=−1<0§u´x=1´¼ê4:§4Š•y=2¶x=−1´¼ê4Œ:§4ŒŠ•y=2.03003(4)Ïy=3sinxcosx(sinx−cosx)=sin2x(sinx−cosx),y=3cos2x(sinx−cosx)+sin2x(cosx+22sinx)πkπ00003√007:•x=kπ+,x=课后答案网(k∈Z)§qy|x=2kπ=−3<0,y|x=2kπ+π=2>0,y|x=2kπ+π=424223√000000−3<0,y|2kπ+π=3>0,y|x=2kπ+5π=−2<0,y|x=2kπ+3π=3>0§422π5πu´x=2kπž§k4ŒŠy=1¶x=2kπ+ž§k4ŒŠy=1¶x=2kπ+ž§k4Œ√242Šy=−¶2√πwww.hackshp.cn23πx=2kπ+ž§k4Šy=¶x=2kπ+πž§k4Šy=−1¶x=2kπ+ž§k4422Šy=−1.x−x0e−e(5)Ïy=1−sinx+sinhx§Ø´¦7:§d−sinx+=0´„x=0´˜‡7:2ππd−sinx,sinhx3−,î‚üN5§•ù7:´•˜.220000000000(4)(4)y=−cosx+coshx,y(0)=0;y=sinx+sinhx,y(0)=0;y=cosx+coshx,y(0)=2>0§u´x=0´¼ê4:§4Š•y=2.000(n−1)(n)6.ef(x)3:xäk†n ëY꧿…f(x)=f(x)=···=f(x)=0,f(x)6=0§@00000(n)(n)on•Ûꞧf(x)š4Š¶n•óêf(x)>0ž§f(x)•4Š¶n•óêf(x)<00000ž§f(x)•4ŒŠ.0000f(x0)2y²µòf(x)3x=x:^VúªÐmµf(x)=f(x)+f(x)(x−x)+(x−x)+···+000020(n)f(x0)nn(x−x)+o((x−x))n!00(n)000(n−1)f(x0)nnÏf(x)=f(x)=···=f(x)=0§f(x)=f(x)+(x−x)+o((x−x))0000n!00若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn87(n)nf(x0)x→xž§o((x−x))→0§x¿©‚Cxž§=|x−x|¿©ž§f(x)−f(x)†(x−00000n!nx)kƒÓÎÒ0(n)ef(x)>0§0(n)nf(x0)n(1)n•Ûꞧex>x§K(x−x)>0§u´(x−x)>0§lf(x)−f(x)>0=f(x)>00n!00f(x0)¶(n)nf(x0)nexx§„´x0§dž(x−0000n!nx)>0(x6=x)§lf(x)−f(x)>0§=3x¿©,•S§ðkf(x)>f(x)§ùL²f(x)´0000004Š.(n)ef(x)<0§0(n)nf(x0)n(1)n•Ûꞧex>x§K(x−x)>0§u´(x−x)<0§lf(x)−f(x)<0=f(x)<00n!00f(x0)¶(n)nf(x0)nex0§lf(x)−f(x)>0=f(x)>f(x0)00n!00Ïdf(x)Ø´4Š.0(n)nf(x0)(2)n•óꞧ•‡x¿©Cx§ØØx>x§„´x0§dž(x−0000n!nx)<0(x6=x)§lf(x)−f(x)<0§=3x¿©,•S§ðkf(x)课后答案网22x−3,x<1Ø•3,x=103¦§y=−2x+3,−1235qy(−10)=132,y(1)=0,ywww.hackshp.cn=,y(2)=0,y(10)=72§¼ê•ŒŠ´132§•Š´0.24ex−3,x>3x−3e,x>30(2)y=3−x§¦§y=Ø•3,x=3§w,Ã7:e,x<33−x−e,x<3828qy(−5)=e,y(3)=1,y(5)=e§¼ê•ŒŠ•e§•Š•1.8.c´þABãål•100úp§ó‚C†Aƒå40úp§ACR†uAB.8‡3AB¥m˜:D•ó‚C?˜^ú´£ã5-21¤§¦løAÕB$Àó‚C¤^$¤•Ž.¯D:AT3Û?º®•z˜úpc´$¤†ú´$¤ƒ"´3:5.)µ|AD|=xúp§K|DB|=100−xúp¶zúpc´$¤•3t§Kzúpú´$¤•5t§o$¤•yt√√√05x−3x2+160000Kyt=x2+1600(5t)+(100−x)(3t)=y=5x2+1600+3(100−x)§u´y=√,y=x2+16008000>0§7:•x=30§…x=30•4:§D:A3åA30úp?.3(x2+1600)29.r˜Š7ç¤Ý/7^.¯Ý/•Ú°õŒž§ ¡Èp•Œº√)µ7 ¡Œ»•R§Ý/•!°©O•x,y§Kx2+y2=2R§u´y=4R2−x2§lS=若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn89y413y2√2033003f(y)=+2p−2py§Kf(y)=(y−2p),f(y)=>0§7:•y=2p§…§Ò´f(y)44p2p2qp2q√233−Š:§Ïd¤¦•áål•d=f(2p)=|p|2+23−24.17.`E±u=20Ô/ž„Ý•ÀÊ1§Ìž3Ù¡h=82Ô?k¯E±v=16Ô/ž„Ý•HÊ1§¯ÛžüEål•Cºp√):xž üEål•C§üEƒåSÔ§KS=(82−16x)2+(20x)2=656x2−2624x+67242000-f(x)=656x−2624x+6724§¦Ù•Š"Kf(x)=1312√x−2624,f(x)=1312>0§7:•x=2…§•f(x)4Š:§K2ž üEål•C§džS=1041.18.²/þ˜˜•Ô§•þ•Pú6.®•ÔN†/¡ ÞXꕵ"y˜åF§¦ÔNm©£Ä.¯då†Y²••Yϕ•õŒž§^å•Žº£ã5-22¤ºµPG)µâK§kFcosϕ=µ(PG−Fsinϕ)=F=cosϕ+µsinϕ-y=cosϕ+µsinϕ§•¦F•§•‡¦y•Œ00000dy=−sinϕ+µcosϕ,y=−cosϕ−µsinϕ§7:•ϕ=arctanµ§džy<0§L²ϕ=arctanµž§y•ŒŠ§lF•Š§=^å•Ž.19.Xã5-23¤«§k`!¯ü)èÜ^˜CØ짯CØìMA3Û?§¤^Ñ>‚•Žº)µMA3†`R†˜ål•xúp?§¤^Ñ>‚l•Ž√2p220xx−3001d®•§l=1+x+2.25+(3−x)§Kl=√+p,l=+222231+x2.25+(3−x)(1+x)22.25>0§7:•x=1.2§…••Š:§=x=1.2úpž§¤^Ñ>‚•Ž.3(2.252+(3−x)2)222xy20.ý+=1ƒ‚†ü‹I¶©OuA,Bü:§a2b2(1)¦ABü:mål•Š¶(2)¦∆OAB•¡È.22bxbx)µƒ:•(x,y)§Kƒ‚Ç•k=−§u´ƒ‚•§•Y−y=−(X−x)§a2ya2y22abØ”˜„5§Œ:M(x.y)31˜–•§ƒ‚3ü‹I¶þ å©O•,§Kxysr4422abab(1)¤¦ABü:mål•d=+=a+x2y2x2a2−x222ab-f(x)=x2+a2−x2§‡课后答案网¦d•Š§•I¦f(x)•Š.222222202a2bx006a2ab+6bx22df(x)=−+,f(x)=+>0§…dux∈[0,a],x6a§K7:x3(a2−x2)2x4(a2−x2)3a3p2÷vx=…džf(x)•Š§=d•Š§•áål•d=af(x)=a+b.a+b231aabab222(2)UK§kS=··√=√§•Ä¼êg(x)=x(a−x)2xa2−www.hackshp.cnx22xa2−x2‡¦S•Š§•‡¦g(x)•ŒŠ0230022a00adg(x)=2ax−4x,g(x)=2a−12x§7:•x=√…džg(x)<0§=x=√žg(x)•22ŒŠ§lS•Š§•¡È•S=ab.αx21.?ؼêx(α>190<α<1),e,lnx,xlnx3(0,+∞)Sà5.α0α−100α−2)µf(x)=x,f(x)=αx,f(x)=α(α−1)x00α00αα>1ž§f(x)>0§Kx3(0,+∞)Seà¶0<α<1ž§f(x)<0§Kx3(0,+∞)Sþà.x0x00xxf(x)=e,f(x)=e,f(x)=e>0(x>0)§Ke3(0,+∞)Seà01001f(x)=lnx,f(x)=,f(x)=−<0(x>0)§Klnx3(0,+∞)Sþàxx20001f(x)=xlnx,f(x)=1+lnx,f(x)=>0(x>0)§Kxlnx3(0,+∞)Seàx22.?Øe¼êà5Ú$:µ23(1)y=3x−x2a(2)y=(a>0)a2+x2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn90(3)y=x+sinx√(4)y=1+x2)µx(−∞,1)(1,+∞)02000000(1)y=6x−3x,y=6−6x§y=0Š•x=1§LXeµyÎÒ+-§$:yeàþà‹I•(1,2)√22202ax002a(3x−a)003(2)y=−,y=§y=0Š•x=±a§LXeµ(a2+x2)2(a2+x2)33√√√√x−∞,−3a−3a,3a3a,+∞√√3333333300§$:‹I•−a,,a,yÎÒ+-+3434yeàþàeà00000(3)y=1+cosx,y=−sinx§y=0Š•x=kπ(k∈Z)§LXeµx((2k−1)π,2kπ)(2kπ,(2k+1)π)((2k+1)π,2(k+1)π)00yÎÒ+-+§$:‹I•(kπ,kπ)yeàþàeà0x00100(4)y=√,y=§Ky>0§¼ê´eà§lÃ$:.2231+x(1+x)2x+123.y²•‚y=kuÓ˜†‚þn‡$:.x2+1√√201−2x−x002(x−1)(x+2−3)(x+2+3)y²µy=,y=(x2+1)2(x2+1)3√√00-y=0§√x1=1,x2=−2+3,x3√=−2−3√√000000x<−2−3ž§y<0¶−2−30¶−2+3−1ž§y>0√√√3+1√1−3u´•‚3x1,x2,x3?kn‡$:A(1,1),B−2+3,,C−(2+3),44√√√131−3−2−331−3LA,B†‚•¤•y=x+§òC:‹I“þ㕧§=+==C÷v444444x+1d•§§K•‚y=kuÓ˜†‚þn‡$:.x2+1024.ef(x)´eà¼ê£½î‚eà¼课后答案网ꤧf(x0)•3§K0f(x)>f(x)+f(x)(x−x)0000(x6=x).f(x)>f(x)+f(x)(x−x)0000y²µx•f(x)½Â•S?˜:§x6=x0£Ø”x>x0¤x+xf(x)−f(x)f(x1)−f(x)x1+x-x0§df(x)•eà¼ê§K000§df(x)•eà¼1=>¶x2=2x−xx1−x200f(x1)−f(x)f(x2)−f(x)x2+xf(x2)−f(x)f(x3)−f(x)ê§K0>0¶x0§df(x)•eà¼ê§K00www.hackshp.cn3=>x−xx2−x2x−xx3−x0000|x−x|f(xn)−f(x)0→0(n→∞)§Kx0Xd?1e§Œê{xn}§|xn−x|=n→x(n→∞)§…>02n0xn−x0f(xn+1)−f(x)0xn+1−x00f(xn)−f(x0)f(xn)−f(x0)0qf(x)•3§Klim=lim=f(x)00n→∞xn−x0xn→x0xn−x0f(x)−f(x0)f(xn)−f(x0)f(x)−f(x0)00q>§Kd4•5Ÿ§>f(x)§lf(x)>f(x)+f(x)(x−x−xx000n−xx−x000x)00ÓnŒy§ef(x)´î‚eà¼ê§Kf(x)>f(x)+f(x)(x−x).00025.ef(x)´eà¼ê§K−f(x)´þà¼ê.x1+x2f(x1)+f(x2)y²µÏf(x)´eà¼ê§Kf(x)3[a,b]þëY§é[a,b]¥?¿ü:x1,x2§ðkf6§22x1+x2f(x1)+f(x2)(−f(x1))+(−f(x2))u´−f>−=§l−f(x)´þà¼ê.22226.(1)efn(x)´eà¼ê§¯F(x)=min{fn(x)}´Ø´eà¼êºn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn91(2)ef(x),g(x)´eà¼ê§¯f(x)+g(x)´Ø´eà¼êº(3)`²ng¼êØ´eà¼ê.)µ(1)ؘ½.~µ1212f1(x)=,f2(x)=x(x>0)ž§f1(x),f2(x)Ñ´eà¼ê§F(x)=min,x3(1,1):Ø÷xxveà¼ê½Â§=F(x)Ø´eà¼ê.2222x2xxf1(x)=x,f2(x)=ž§f1(x),f2(x)Ñ´eà¼ê§…F(x)=minx,=´eà¼ê.222(2)f(x)+g(x)´eà¼ê.x1+x2f(x1)+f(x2)Ïf(x),g(x)´eà¼ê§Kf6,22x1+x2g(x1)+g(x2)x1+x2x1+x2x1+x2f(x1)+f(x2)g6§u´(f+g)=f+g6+222222g(x1)+g(x2)1=[(f+g)(x1)+(f+g)(x2)]=f(x)+g(x)´eà¼ê.22320200(3)f(x)=ax+bx+cx+d(a6=0)§Kf(x)=3ax+2bx+c,f(x)=6ax+2bu´§b00b00a>0ž§x>−ž§f(x)>0§f(x)´eà¼ê¶x<−ž§f(x)<0§f(x)´þà¼ê3a3ab00b00a<0ž§x>−ž§f(x)<0§f(x)´þà¼ê¶x<−ž§f(x)>0§f(x)´eà¼ê3a3abKf(x)Ø´eà¼ê§3x=−?k$:.3ah−h2x227.XÛÀJëêh>0§•U¦•‚y=√e3x=±σ(σ>0,󕮉½~ê)?k$:.π3302h−h2x2002h−h2x222)µy=−√xe,y=√e(2hx−1)ππ001-y=0§Kx=±√2h1001100100x<−√ž§y>0§•‚eà¶−√√ž§y>2h2h2h2h0§•‚eà课后答案网111K3x=±√?kü‡$:§u´±√=±σ§qh,σ>0§Kh=√.2h2h2σ2x28.¦y=4Š9$:§¿¦$:?ƒ‚•§.x2+1202x002−6x)µy=,y=(1+x2)2(1+x2)3xwww.hackshp.cn(−∞,0)0(0,+∞)07:•x=0§ÑeLµyÎÒ-0+y&4Š0%Kx=0§yk4√Šy=0.003-y=0§Kx=±§ÑeLµ√3√√√3333x−∞,−−,,+∞333300yÎÒ-+-yþàeàþà√√3131$:•−,,,.3434√√−23√311333$:−,?ƒ‚•§•y−=2x+34413+13√=33x+8y+1=0¶若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn93(iv)x→∞ž§y→0§y=0´•‚˜^Y²ìC‚.y6√-3-1√-x013(3)(i)½Â•(−∞,+∞)§´ó¼ê§•‚"uy¶é¡.0−x200−x220100(ii)y=−10xe,y=10e(2x−1)§x=0ž§y=0¶x=±√ž§y=0.2(iii)L?ØXeµ1√1111x(−∞,−√)−DF12(−√,0)0(0,)√(√,+∞)22sqrt222y0+++0---y00+0---0+55yeà%√þà%4ŒŠþà&√eà&ee5(iv)x→∞ž§y→0§y=0´•‚˜^Y²ìC‚.y65-101x-√√课后答案网22x−xe+e(4)(i)½Â•(−∞,+∞)§´ó¼ê§•‚"uy¶é¡£ù´V•{u¼êcoshx=¤.20000(ii)y=sinhx,y=coshx§x=0ž§y=0¶0000duy>0(x∈(−∞,+∞)§y3(−∞,+∞)þ´eà.Ïy|x=0=1>0§ymin=1.ywww.hackshp.cn61-xSS(5)(i)½Â•(−∞,−1)(−1,1)(1,+∞)§´ó¼ê§•‚"uy¶é¡.202x002(x+1)00(ii)y=−,y=§x=0ž§y=0¶x=±1ž§yØ•3¶x=(x2−1)2(x2−1)300±1ž§yØ•3.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn95√!√√!14+614+614+6x1.4,22(2,+∞)1010,y0+++0+y00+0-0+yeà%-0.0186þà%0eà%y612-0x-8S(8)(i)½Â•(−∞,−1)(−1,+∞).206(x−1)0012(x−1)(x−3)000(ii)y=,y=−§x=1ž§y=0¶x=1,x=3ž§y=0¶(x+1)4(x+1)5000x=−1ž§y,yþØ课后答案网•3.(iii)L?ØXeµx(−∞,−1)-1(−1,1)1(1,3)3(3,+∞)y0+Ø•3+0+++y00-Ø•3-0+0-1yþà%ýÂþà%0eà%þà%8−(iv)x→−1ž§y→+∞§x=−1´•‚˜^R†ìC‚¶x→∞ž§y→1§www.hackshp.cny=1´•‚˜^Y²ìC‚.y6--1013x(9)(i)½Â•(−∞,+∞).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn97(3)L?ØXeµx(−∞,x1)x1(x1,−1)-1(−1,0)0y0---Ø•3-0y00+0-ýÂ--yeà&$:þà&ýÂþà&4Š9x(0,1)1(1,3)3(3,+∞)y0+Ø•3+0-y00-Ø•3---yþà%ýÂþà%4ŒŠþà&9−2(4)x→±1ž§y→∞§x=±1´•‚R†ìC‚.y60--27x课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn98§4.²¡•‚•Ç21.¦•‚y=4x−x•Ç±93:(2,4)•ÇŒ».002000y2)µÏy=4x−x§y=4−2x,y=−2§K•ÇK=3=3§u´•ÇŒ(1+y02)2[1+4(2−x)2]211231»ρ==[1+4(x−2)]2§l3:(2,4)•ÇŒ»ρ=.K222.¦e•‚•Ç†•ÇŒ»µx(1)]ó‚y=acosh(a>0)a2(2)Ô‚y=2px(p>0)(3)^Ó‚x=a(t−sint),y=a(1−cost)(a>0)(4)%9‚ρ=a(1+cosθ)(a>0)22(5)VÝ‚ρ=2acos2θ(a>0)λθ(6)éêÚ‚ρ=ae(λ>0))µ00x0x001xy1(1)Ïy=acosh§y=sinh,y=cosh§K•ÇK=3=x§u´•ÇŒaaaa(1+y02)2acosh2a12x»ρ=acosh.Ka2002200p00p0pyp(2)Ïy=2px§K2yy=2p=y=§y=−2y=−3§K•ÇK=3=3=yyy(1+y02)2(y2+p2)233!p2112x2(y2+p2)2=§u´•ÇŒ»ρ==p1+½.332(2px+p2)22x2Kppp1+p000000(3)Ïx=a(t−sint),y=a(1−cost)§x=a(1−cost),x=asint;y=asinty=acost§K•x0y00−x00y011tpÇK=3=§u´•ÇŒ»ρ==4asin(=22ay).(x02+y02)2tK24asin2课后答案网20200000ρ+2ρ−ρρ3(4)Ïρ=a(1+cosθ)§ρ=−asinθ,ρ=−acosθ§K•ÇK=3=√§u´•(ρ2+ρ02)222aρ√22aρÇŒ»R=.32442020022020−2asin2θ004a+ρρ+2ρ−ρρ(5)Ïρ=2acos2θ§K2ρρ=−4asin2θ§ρ=,ρ=−3§K•ÇK=3=www.hackshp.cnρρ(ρ2+ρ02)223ρ2a§u´•ÇŒ»R=.2a23ρ20200λθ0λθ002λθ2ρ+2ρ−ρρ1(6)Ïρ=ae§ρ=λae=λρ,ρ=aλe=λρ§K•ÇK=3=1§(ρ2+ρ02)2|ρ|(1+λ2)2√u´•ÇŒ»R=|ρ|1+λ2.23.¦•‚y=2(x−1)••ÇŒ».0232320001(1+y)2[1+16(x−1)]2)µÏy=2(x−1)§y=4(x−1),y=4§K•ÇŒ»R===Ky004231‡¦R•§K7k[1+16(x−1)]2•§=x=1ž§Rmin=.42x4.˜œÅ÷Ô‚´»y=£ü•’¤Š:Àœ1§3‹I:O„Ýv=140’/¦§œ1N4000•G=70ú6.¦džŒ«éœ1‡å.2mv)µdÔnÆ•£•§Š!„±$ÄÔN¤É•%å•F=§Ù¥m•ÔNŸþ§v•§„R若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn99ݧR•Œ».2mv¤¦Œ«éœ1‡åŒA•F=Gg+§Ù••A••%.R2230x0011(2000+x)2âK¿§k¦•ÇŒ»§y=,y=§K•ÇŒ»R==§u´3‹I20002000K20002:OR=2000£’¤§q3‹I:O„Ýv=140’/¦§lF=1372(N).5.˜å••þ´P§± „v¨Lÿx£ã5-32¤§x¡ACB´˜Ô‚§Ùº€Xã«.¦ð•LC:žéx¡Øå.4δ2)µ±O•:§AB•x¶§CO•y¶ïá‹IX§KÔ‚•§y=−x+δl22mvdÔnÆ•§ð•LC:žéx¡Øå•F=cosθ+mgR2308δ008δ1(l+8δx)2âK¿§k¦•ÇŒ»§y=−x,y=−§K•ÇŒ»R==§u´3:CR=l2l2K8lδ2222lPvgl−8δv§q3:Cθ=π§lF=Pg+cosθ=P.8δRl2课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn100§5.–½.1.|^â7ˆ{K¦e4•µtanax(1)limx→0sinbx21−cosx(2)limx→0x3sinxπ−arctanx(3)lim2x→∞1sinxbx(4)limx→∞eax11(5)lim−x→1lnxx−1x(6)lim(π−x)tanx→π2ln(cosax)(7)limx→0ln(cosbx)cos(sinx)−cosx(8)limx→0x4xxa−b(9)limx→0xx−1(10)limx→1lnxxaa−x(11)lim(a>0)x→ax−a1−2sinx(12)limx→πcos3x6lnx(13)limx→0cotxclnx课后答案网(14)lim(b,c>0)x→+∞xb1(1+x)x−e(15)limx→0xbc(16)limxlnx(b,c>0)x→0sinx(17)limxx→0www.hackshp.cn1(18)limx1−xx→111(19)lim−x→0xex−1x1(20)limlnx→+0x)µ2tanaxasecaxa(1)lim=lim=x→0sinbxx→0bcosbxb22221−cosx1−cosx2xsinxsinx1(2)lim=lim=limlim=x→0x3sinxx→0x4x→04x3x→02x22π1−arctanx−221+x2x(3)lim=lim=lim=1x→∞1x→∞11x→∞1sin−cos(1+x2)cosxx2xx若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn102ln1xlimxln1x→+0x(20)limln=ex→+0xx11ln(lny)11-y=§Klimxln(ln)=lim=lim=0§llimln=1xx→+0xy→+∞yy→+∞ylnyx→+0x2.Á`²e¼êØU^â7ˆ{K¦4•µ21xsin(1)limxx→0sinxx+sinx(2)limx→∞x−cosx2x+sin2x(3)limx→∞(2x+sinx)esinx2(x−1)sinx(4)limx→1πln1+sinx2)µ21111xsin2xsin−coscos(1)Ïx©f!©1Óžéx¦ê§xx§xx→0ž4•Ø•3§Ïdâsinxcosxcosx21xsinxx17ˆ{KØU·^§´4•´•3"¯¢þ§klim=lim·xsin=0x→0sinxx→0sinxxx+sinx1+cosx(2)Ï©f!©1Óžéx¦ê§§x→∞žd¼ê4•Ø•3§Ïdâ7ˆ{x−cosx1+sinxsinx1+x+sinxxKØU·^§´4•´•3"¯¢þ§klim=limcosx=1x→∞x−cosxx→∞1−x0π001(3)éuØÓSµxn=2nπ+9xn=2nπ(n=1,2,···)§n→∞ž§KØÓ4•91§l2e4•Ø•3.2x+sin2x^â7ˆ{K¦)§klim=x→∞(2x+sinx)esinx2+2cos2xlim=x→∞(2+cosx+2xcosx+sinxcosx)esinx24cosx课后答案网lim=x→∞[2+cosx(1+2x+sinx)]esinx4sinx−121sinxlim§Ïe>e,1+2x+sinx>2x§K[2+(1+2x+sinx)]e>x→∞21cosxcosx+(1+2x+sinx)esinxcos2xcosx−12x+sin2xe(−2+2|x|)→+∞(x→∞)§Klim=0.x→∞(2x+sinx)esinx(x2−1)sinwww.hackshp.cnx(4)†¦4•Œlim=0§d4•ØÎÜ^â7ˆ{K¦4•^‡.x→1πln1+sinx2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn103§6.•§Cq)31.¦•§x−x−4=0Š§¦Ø؇L0.0001.302)µf(x)=x−x−4§3[1,2]m§f(1)=−4<0,f(2)=2>0=f(1)f(2)<0…f(x)=3x−1>000,f(x)=6x>000Ïf(2)f(2)=24>0§Kl:(2,f(2))=:(2,2)m©Šƒ‚§x=2ŠÐŠ.0f(2)f(x1)f(x2)f(x3)u´x1=2−≈1.81818,x2=x1−≈1.79663,x3=x2−≈1.79632,x4=x3−≈f0(2)f0(x1)f0(x2)f0(x3)1.79632x3†x4c5êƒÓ§ùL«®CuŠ°(Š"•`²°(ݧ^1.7963Á˜e§kf(1.7963)≈−0.00019<0§f(1.79632)≈0.00002>0§e1.7963Š•ŠCqŠ§KØ؇L0.0001.32.¦•§x−x−4=0Š§¦Ø؇L0.0001.32020)µf(x)=x−5x+6x−1§f(0)=−1<0,f(1)=1>0,f(x)=3x−10x+6§džf(0)=6>√√√005−705−705−70,f(1)=−1<0§3(0,1)Sf(x)k":§džf(x)30,S•¶f(x)3,1S333•K.y©O•Äf(x)3(0,0.7)†(0.7,1)¥ŠÏf(0.7)=1.093>0§3(0,0.7)¥7k¢Šξ§3(0.7,1)¥ÃŠ.0000y¦ξ,f(x)=6x−10<0(∀x∈(0,0.7))§Ïf(0)=−1,f(0)=−10§x=0ŠÐŠ.0f(0)f(x1)f(x2)f(x3)u´x1=0−≈0.16667,x2=x1−≈0.19706,x3=x2−≈0.19806,x4=x3−≈f0(0)f0(x1)f0(x2)f0(x3)0.19806x3†x4c5êƒÓ§ùL«®CuŠξ°(Š"•`²°(ݧ^0.1980Á˜e§kf(0.1980)≈−0.00026<0§f(0.1981)≈0.01397>0§e0.1980Š•ŠCqŠ§KØ؇L0.0001.课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn1041Ü©üCþÈ©Æ18Ùؽȩ§1.ؽȩVg9${KRR11.y²µef(t)dt=F(t)+C§Kf(ax+b)dx=F(ax+b)+C.aR00110y²µÏf(t)dt=F(t)+C§[F(t)+C]=f(t)§KT(ax+b)=[F(ax+b)]=f(ax+b)§uaaR1´f(ax+b)dx=F(ax+b)+C.a2.¦eؽȩµZ2(1)(2−secx)dxZ√43x(2)x−2x+dx2Z√√22(3)x+3x+√+√−2dxx3xZx111(4)e+++dxxx2x3Z1(5)2cosx+sinxdx2Z21(6)cosx−+√dx1+x241−x2Z1(7)cosx+sinx+1dx2Zxxx1e(8)2+−dx35Z23(9)(3−x)dxZq1√(10)1−xxdxx2课后答案网)µZ2(1)(2−secx)dx=2x−tanx+CZ√43x151413(2)x−2x+dx=x−x+x2+C2523Z√√22233421(3)x+3x+√+√−2dx=x2+x3−2x+3x3+4x2+Cx3xwww.hackshp.cn34Zx111x11(4)e+++dx=e+ln|x|−−+Cxx2x3x2x2Z11(5)2cosx+sinxdx=2sinx−cosx+C22Z211(6)cosx−+√dx=sinx−2arctanx+arcsinx+C1+x241−x24Z11(7)cosx+sinx+1dx=sinx−cosx+x+C22Zxxxxx1e1x11e(8)2+−dx=2−−+C35ln2ln335ZZ2324639517(9)(3−x)dx=(27−27x+9x−x)dx=27x−9x+x−x+C57ZqZ1√3−547−1(10)1−xxdx=(x4−x4)dx=x4+4x4+Cx27若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn106Z2x+7(25)dxx2−2x−3Z2x−1(26)dxx4+1)µZZdx1d(5x−7)1(1)==ln|5x−7|+C5x−755x−75ZZ11(2)cos(ωt−ϕ)dt=cos(ωt−ϕ)d(ωt−ϕ)=sin(ωt−ϕ)+CωωZZdx+3dx2x(3)r2=2r2=2arcsin2+3+Cxx1−+31−+322Z√Z√√dx2d(x)2√(4)√=q√=arcsin(2x)+C1−2x221−(2x)22ZZ10210111(5)tanxsecxdx=tanxd(tanx)=−tanx+C11ZZαxαxxαx(2e)(6)e·2dx=(2e)dx=+Cln(2eα)ZZxxxx2xxx42x9(7)(2+3)dx=(4+2·6+9)dx=+6++Cln4ln6ln9ZZZsinxd(cosx)(8)tanxdx=dx=−=−ln|cosx|+C=ln|secx|+CcosxcosxZpxdxZppp(9)tan1+x2·√=tan1+x2d(1+x2)=ln|sec1+x2|+C1+x2ZZ2µ+12µ12µ2(αx+β)(10)(αx+β)xdx=(αx+β)d(αx+β)=+C2α2α(µ+1)ZZdx2xxx(11)=cscd=−cot+C1−cosx222ZZdx11Atanx1A(12)A2sin2x+B2cos2x=AB2dB=ABarctanBtanx+CA1+tan2x课后答案网BZZsinx·cosx11212(13)dx=d(sinx)=arctan(sinx)+C1+sin4x21+(sin2x)22ZZdx2πππ(14)=cscx+dx+=−cotx++Csin2x+π4444ZpZp2831833839(15)x1+xdx=1+www.hackshp.cnxd(1+x)=(1+x)8+C327ZZ23sinxcosx1d(1+sinx)13(16)dx==ln(1+sinx)+C1+sin3x31+sin3x3ZZZ1−2sinx2dcosx(17)dx=secxdx+2=tanx−2secx+Ccos2xcos2xZZxdxdex(18)==arctan(e)+Cex+e−xe2x+1ZZsinx+cosxd(sinx−cosx)32(19)√dx=√=(sinx−cosx)3+C3sinx−cosx3sinx−cosx2ZZZ21+sin2x2d(sinx)2(20)dx=cscxdx+=−cotx+ln(sinx)+C=−cotx+2ln|sinx|+Csin2xsin2xsZ√ln(x+1+x2)(21)dx=1+x2Zqppp23ln(x+1+x2)d(ln(x+1+x2)=[ln(x+1+x2]2+C3若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn107ZZdxde−xp(22)√=−√=−ln(e−x+1+e−2x)+C1+e2x1+e−2xZZdxd(x−1)(23)==arctan(x−1)+Cx2−2x+2(x−1)2+1ZZdxd(arcsinx)1(24)√==−+C(arcsinx)21−x2(arcsinx)2arcsinxZZ2x+72x+10(25)dx=1+dx=x2−2x−3(x+1)(x−3)Z224(x−3)1−+dx=x−2ln|x+1|+4ln|x−3|+C=x+2ln+Cx+1x−3|x+1|1ZZZdx+2−2x−11−xx(26)x4+1dx=x2+x−2dx=2=1x+−2x√√x+1−2√√22x2−2x+1lnx√+C=ln√+C4x+1+24x2+2x+1x2.¦eؽȩµZ√sinx(1)√dxxZ√2(2u+1)(2)duu2Z√x+1(3)edxZ2x(4)√dx4−x2Zp(5)x2+a2dxZp(6)x2−a2dxZ课后答案网dx(7)p(x−a)(b−x)Zdx(8)px2αx2+βZxdx(9)√5+x−x2www.hackshp.cnZp(10)2+x−x2dx)µZ√Zsinx√√√(1)√dx=−2sinxd(x)=−2cosx+CxZ√Z2(2u+1)441−11(2)du=++du=4ln|u|−8u2−+Cu2u3u2uu2ZZ√√√2x+1tt(3)-1+x=t§Kx=t−1,dx=2tdt§u´edx=2tedt=2(t−1)e+C=2(1+x−√1+x1)e+CZZZZx24−x2−4pdxxpx(4)√dx=−√dx=−4−x2dx+4√=−4−x2−2arcsin+4−x24−x2p4−x222xxx4arcsin+C=2arcsin−4−x2+C222若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn108ZZZZppx2ppa2(5)I=x2+a2dx=xx2+a2−√dx=xx2+a2−x2+a2dx+√dx=ppx2+a2x2+a2xx2+a2−I+a2ln|x+x2+a2|+C1§√√xpa2pu´2I=xx2+a2+a2ln|x+x2+a2|+C22ln(x+221§lI=x+a+x+a)+C(C=22C1)2ZZZZppx2ppa2(6)I=x2−a2dx=xx2−a2−√dx=xx2−a2−x2−a2dx−√dx=ppx2−a2x2−a2xx2−a2−I−a2ln|x+x2−a2|+C1§√√xpa2pu´2I=xx2−a2−a2ln|x+x2−a2|+C22221§lI=x−a−ln(x+x−a)+C(C=22C1)2ZZdxdx(7)p=p=(x−a)(b−x)−[x2−(a+b)x]−abZda+ba+bx−x−222x−a−bs=arcsin+C=arcsin+C£Ù¥a0)ax2+bx+cZxdx(12)p课后答案网4x3(a−x)Zp(13)xx4+2x2−1dxZp(14)2+x−x2dxZ2√xdxwww.hackshp.cn(15)1+x−x2Z2x+1(16)√dxxx4+1Z6(17)sinxdxZ24(18)sinxcosxdxZ44(19)sinxcosxdxZ4cosx(20)dxsin3xZdx(21)sin3xcos5xZ(22)tanx·tan(x+a)dx若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn112Z(23)sin5xcosxdxZ2sinx(24)dx1+sin2xZdx(25)sin(x+a)sin(x+b)Zx(26)xecosxdxZdx(27)(2+cosx)sinxZp(28)ln(x+1+x2)2dxZsinxcosx(29)dxsinx+cosxZlnx(30)dx3(1+x2)2Zx(31)xesinxdxZ3xarccosx(32)√dx1−x2Z2(33)(x+|x|)dxZ2x(34)xecosxdxZxxe(35)dx(1+x)2Z√2(36)xlnxdxZdx(37)p(x−a)(b−x)Z1+x(38)xlndx1−xZ课后答案网2(39)xarctanx·ln(1+x)dxZ(40)sinh2xcosh2xdx)µ2x1−twww.hackshp.cn2dt(1)-tan=t§Kcosx=,dx=§21+t21+t2ZZ3+tanxdx213+t12u´4+5cosx=(3−t)(3+t)dt=3ln3−t+C=3lnx+C3−tan2x2t2t2dt(2)-tan=t§Ksinx=,tanx=,dx=§Z2Z1+t21−t21+t2dx1−t21t21x1x2u´=dt=ln|t|−+C=lntan−tan+Csinx+tanx2t242242ZZZx−1Zs2xdxxdx21dx211(3)√=s=s+s=−−x−+5+x−x2211221122211242−x−dx−x−−x−42424211x−2p12x−1arcsin√+C=−5+x−x2+arcsin√+C2212212若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn113√√3(4)-t=41+x4§Kx=4t4−1,dx=t3(t4−1)−4dtZZZZ1t2111111t−11u´√dx==−dt+dt=ln+arctant+x41+x4t4−14t−1t+121+t24t+12√141+x4−11pC=ln√+arctan(41+x4)+C441+x4+12ZZZxdx1√1√11x√13(5)√=xd4x+2=x2+4x−(2+4x)2dx=2+4x−(2+4x)2+C2+4x222212ZZcosxdsinx(6)dx==ln(1+sinx)+C1+sinx1+sinx√(7)-6x=t§Kx=t6,dx=6t5dtZdxu´√√=x(1+2x+3x)ZZdt116t−16=6−−dtt(1+2t3+t2)t4(t+1)4(2t2−t+1)ZZZ26t−13d(2t−t+1)1dt3214t−1qdt=+=ln|2t−t+1|+√arctan√+4(2t2−t+1)82t2−t+182t2−t+18477C1§Zdx39234t−1√6l√√=6ln|t|−ln|t+1|−ln|2t−t+1|−√arctan√+C=6ln|x|−x(1+2x+3x)24√277√3√9√√346x−13x3xln|6x+1|−ln|23x−6x+1|−√arctan√+C=ln√√√−244(1+6x)2(23x−6x+1)3√277346x−1√arctan√+C277Z√√Z√√Zx+1−x−1(x+1−x−1)2px2(8)√√dx=√√√√dx=(x−x2−1)dx=−x+1+x−1(x+1+x−1)(x+1−x−1)2xp1px2−1+ln|x+x2−1|+C22r32x+1t+16t(9)-3=t§Kx=,dx=−dtx−1t3−1(t3−1)2ZZrdx3333x+1u´p=−dt=−t+C=−+C3(x+1)2(x−1)4222x−1√(10)-4x=t§Kx=t4,dx=4t3dtZZZdxt1142u´√√=4dt=4−dt=−++C=x(1+4x)3(1+t)3(1+t)2(1+t)31+t(1+t)24√2√课后答案网−++C1+4x(1+4x)2√bZZdax+dx12a(11)√2=√a22=ax+bx+c√b4ac−bax++2a4a1√bp√lnax++ax2www.hackshp.cn+bx+c+Ca2ara−x(12)-4=txZZ2xdx4atu´p=−dt=4x3(a−x)(1+t4)2Z2ZZZtadtadtdt−4a√√dt=−√−√+a(t2+2t+1)(t2−2t+1)2(t2−2t+1)22(t2+2t+1)2t4+1√2ZZdt−√dt22t−2√√q2√2="√2#2=2√+2arctan(2t−1)+C1(t−2t+1)212(t−2t+1)t−+22Z√dt2t+2√√√=√+2arctan(2t+1)+C2(t2+2t+1)22(t2+2t+1)ZZZ2211t+11t−1dt=dt−dtt4+12t4+12t4+1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn11411ZZ1+Zdt−22t+1t2t1t−1t4+1dt=1dt=2=√arctan√+C3,t2+t−1+222tt2tZ√22t−11t−2t+1dt=√ln√+C4t4+122t2+2t+1Z√√√xdxat3a22tat2+2t+1at2−1lp=−−arctan+√ln√+√arctan√+C§4x3(a−x)1+t421−t242t2−2t+1222tra−xÙ¥t=4.xZZp1px2+1p1p(13)xx4+2x2−1dx=(x2+1)2−2dx2=x4+2x2−1−ln(x2+1+x4+2x2−1)+242CsZpZ9122x−1p92x−1(14)2+x−x2dx=−x−dx=2+x−x2+arcsin+C42483ZZZx2dxpx+12x−1p52x−1(15)√=−1+x−x2dx+√dx=−1+x−x2−arcsin√−1+x−x21+x−x2485p32x−12x+3p72x−11+x−x2+arcsin√+C=−1+x−x2+arcsin√+C25485ZZ1+1Zdx−1r!√x2+12x11x2−1+x4+1(16)√dx=rxdx=s=lnx−+x2++C=ln+xx4+112xx2xx2+x−1+2x2xCZZ3Z61−cos2x12313(17)sinxdx=dx=(1−3cos2x+3cos2x−cos2x)dx=x−sin2x+Z2Z88163121333113(1+cos4x)dx−cos2xdsin2x=x−sin2x+x+sin4x−sin2x+sin2x+16168161664164851313C=x−sin2x+sin4x+sin2x+CZ16464Z48241515(18)sinxcosxdx=−sinxdcosx=−sinxcosx+Z55161515131311cosx=−sinxcosx+x−sin2x+sin4x+sin2x+C=x−sin2x+555164644816203131课后答案网5sin4x+sin2x−sinxcosx+C3202405ZZ4Z2Z44sin2x11−cos4x12(19)sinxcosxdx=dx=dx=(1−2cos4x+cos4x)dx=2162643sin4x1x−+sin8x+C1281281024ZZZZ4323cosx1311cosx3cosxcosx3dx(20)dx=−cosxd=−·−dx=−−+sin3x2www.hackshp.cnsin2x2sin2x2sinx2sin2x2sinxZZsec2x333cosx32x3cosx3x3sinxdx=−−d−cosx=−−lntan−cosx+C22sin2x2x222sin2x222tan2ZZZZZ2222dxsinx+cosxdxdxsinx+cosx(21)=dx=+=dx+sin3xcos5xsin3xcos5xsinxcos5xsin3xcos3xsinxcos5xZZZZZ2222sinx+cosxsinxdxdx1−4sinx+cosxdx=dx+2+=cosx+2dx+sin3xcos3xcos5xsinxcos3xsin3xcosx4sinxcos3xZZZZ22sinx+cosx14sinxdxcosx142dx=secx+2dx+3+dx=secx+secx+sin3xcosx4cos3xsinxcosxsin3x4Zdtanx12142121432123−cscx=secx+secx+3ln|tanx|−cscx+C1=tanx+tanx−cotx+tanx2424223ln|tanx|+CZZZZ22tanx+tanatanx+tanxtana+1−11+tanx(22)tanx·tan(x+a)dx=tanx·dx=dx=dx−ZZ1−tanxtana1−tanxtana1−tanxtanadtanxcosxdx=−x=−cotaln|1−tanxtana|−x+C1=cotaln−x+C1−tanxtanacos(x+a)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn115ZZ111(23)sin5xcosxdx=(sin6x+sin4x)dx=−cos6x−cos4x+C2128ZZZZ√√22sinx1cscxdcotx22(24)dx=dx=1−dx=x+=x+arctancotx+1+sin2xcsc2x+11+csc2x2+cot2x22C(25)sin(a−b)6=0§ZZZdx1sin[(x+a)−(x+b)]1cos(x+b)cos(x+a)K=dx=−dx=sin(x+a)sin(x+b)sin(a−b)sin(x+a)sin(x+b)sin(a−b)sin(x+b)sin(x+a)1sin(x+b)ln+Csin(a−b)sin(x+a)ZZZZxxxxxx(26)I=xecosxdx=xecosx−e(cosx−xsinx)dx=xecosx−ecosxdx+xesinxdx=Zxsinx+cosxxxxxsinx+cosxxxxecosx−e+xesinx−e(sinx+xcosx)dx=xecosx−e+xesinx−2Z2sinx−cosxxxxe−xecosxdx+C1=e(xcosx+xsinx−sinx)−I+C1§2ZxxeKI=xecosxdx=(xcosx+xsinx−sinx)+C22x2t1−t2dt(27)-tan=t§Ksinx=,cosx=,dx=Z21+Zt21+tZ21+t22dx1+t12t121x2xu´=dt=+dt=ln|t(t+3)|+C=lntantan+3+(2+cosx)sinxt(3+t2)3t3(3+t2)3322C=21(1−cosx)(2+cosx)ln+C6(1+cosx)3Zpp(28)ln(x+1+x2)2dx=xln(x+1+x2)2−ZZ1pxpd(1+x2)x·√·2(x+1+x2)·1+√dx=xln(x+1+x2)2−√=(x+1+x2)21+x21+x2ppxln(x+1+x2)2−21+x2+CZZsin2x+π−1√ZZ√sinxcosx42dx=2π1dx=−2π(29)sinx+cosxdx=√π2sinx+4dx−√π2cosx+4−2sinx+22sinx+44Zdtanx+π√128=12xπ√xπ2(sinx−cosx)−4lntan2+8+C22tan+课后答案网28ZZZlnxxxlnxdxxlnxp(30)dx=lnxd√=√−√=√−ln(x+1+x2)+C231+x21+x21+x21+x2(1+x)2ZZZZxxxxxx(31)xesinxdx=xesinx−e(sinx+xcosx)dx=xesinx−esinxdx−xecosxdx=xxxsinx−cosxxeexesinx−e−(xcosx+xsinx−sinx)+C=(xsinx−xcosx+cosx)+CZ2www.hackshp.cn2Z2Zx3arccosxppp(32)√dx=−x2arccosx1−x2+2xarccosx1−x2dx−x2dx=−x21−x2arccosx−1−x2Z3p3x2232222x223−(1−x)2arccosx−(1−x)dx=−x1−xarccosx−−(1−x)2arccosx−333332x36x+x32+x2px−+C=−−1−x2arccosx+C3393ZZZ222323232322(33)(x+|x|)dx=(2x+2x|x|)dx=x+2x·sgnx·xdx=x+xsgnx+C=x+x|x|+C33333ZZ2x2xx22xx(34)xecosxdx=xecosx−e(2xcosx−xsinx)dx=xecosx−e(xcosx+xsinx−sinx)+2xxZesinx−x22xxxe(2xsinx+xcosx)dx=xe(cosx+sinx)−e(xcosx+xsinx−sinx)−e(xsinx−xcosx+Z2xx22cosx)−xecosxdx+C1=e(xcosx+xsinx−2xsinx+sinx−cosx)−I+C1§Zx2xe22KI=xecosxdx=(xcosx+xsinx−2xsinx+sinx−cosx)+C2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn116ZZZxxxxxex1xexxexe(35)dx=−xed=−+edx=−+e+C=+C(1+x)21+x1+x1+xx+1ZZZ√222341223838√223(36)xlnxdx=lnx·x2−x2lnxdx=lnx·x2−x2lnx+xdx=lnx·x2−33399383163232x2lnx+x2+C=x2(9lnx−12lnx+8)+C92727r2b−xb+atb−a2(b−a)t(37)-t=§Kx=,x−a=,dx=−dt§x−a1+t21+t2(1+t2)2ZZrdxdtb−xu´p=−2=−2arctant+C=−2arctan+C(x−a)(b−x)1+t2x−aZZZZ2221+xx1+xxx1+xdx121+x(38)xlndx=ln−dx=ln−+dx=xln−1−x21−x1−x221−x1−x221−x11+xln+x+C21−xZ22x2(39)xarctanx·ln(1+x)dx=arctanxln(1+x)−Z2ZZ22212ln(1+x)2xarctanxx2121ln(1+x)x+dx=arctanxln(1+x)−ln(1+x)dx+dx−21+x21+x22221+x2ZZZ22xarctanxx2x2x1xarctanxdx+dx=arctanxln(1+x)−ln(1+x)+dx+arctanxln(1+1+x2221+x22ZZZ2222xarctanxxarctanxx1xx2x)−dx+dx−arctanx+dx=arctanxln(1+x)−1+x21+x2221+x222x23312x122ln(1+x)+x−arctanx+arctanxln(1+x)−arctanx+C=arctanx[xln(1+x)+22222222x23ln(1+x)−x−3]−ln(1+x)+x+C22ZZZ221211x(40)sinhxcoshxdx=sinh2xdx=(cosh4x−1)dx=sinh4x−+C48328课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn1171Ôٽȩ§1½È©Vg|^½È©½ÂOÈ©µZl(1)f(x)dx§Ù¥f(x)=ax+b,a,b´~ê0Z22(2)xdx−1Z1x(3)adx0)µl(1)Ïf(x)3[0,l]þëY§½È©73§â½È©½Â§ò«m[0,l]n ©§Kzf«m∆xi=§niξizf«m[xi−1,xi]mà:§=ξi=l(i=1,2,···,n).nnnnnXXXialXln+12È©Úf(ξi)∆xi=(aξi+b)∆xi=l+b=(nb+ia)=bl+al§nnn22ni=1i=1i=1i=1ZlXnn+12a2u´f(x)dx=lim(aξi+b)∆xi=limbl+al=bl+l0||x||=1→0n→∞2n2ni=12(2)Ïx3[−1,2]þëY§½È©73§â½È©½Â§ò«m[−1,2]n ©§Kzf«m∆xi=33i3i−n§ξizf«m[xi−1,xi]mà:§=ξi=−1+=(i=1,2,···,n).nnnn2nnnXX2X3i−n3X322911È©Úf(ξi)∆xi=ξi∆xi==(9i−6ni+n)=1+2+−nnn32nni=1i=1i=1i=1191++3§nZ2Xn229111u´xdx=limξi∆xi=lim1+2+−91++3=3−1||x||=1→0n→∞2nnnni=1x1(3)Ïa3[0,1]þëY§½È©73§â½È©½Â§ò«m[0,1]n ©§Kzf«m∆xi=§niξizf«m[xi−1,xi]mà:§=ξi=(i=1,2,···,n).课后答案网n1XnXnXnan(1−a)Z1Xnξi1i1,a6=1xξiÈ©Úf(ξi)∆xi=a∆xi=an=n(1−an)u´adx=lima∆xi=n0||x||=1→0i=1i=1i=11,a=1ni=11an(1−a)a−1lim=,a6=11n→∞n(1−an)lna1,a=1www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn118§2½È©3^1.äe¼êÈ5µ1(1)f(x)3[−2,2]þk.§ØëY:´(n=1,2,3,···)nπ(2)f(x)=sgnsin§3[0,1].x)(1)f(x)3[−2,2]þ´È.Ïf(x)3[−2,2]þk.§∃M>0§¦|f(x)|6M,∀x∈[−2,2]§lÙÌω(f)62M.2M11∀ε>0§g,êN÷vN=+1§u´3,2þf(x)kkõØëY:§Ïf(x)3,2þεNNÈ.Xn130,þ§òÙ©Ü©«m∆xi§1i«m∆xiþÌωi(f)6ω(f)62M§Kωi(f)∆xi6Ni=1Xn2Mε12M∆xi6<2M·=ε§f(x)30,þ´È.N2MNi=1qduf(x)3[−2,0]þëY§,È.âÈ©"u«m5§f(x)3[−2,2]þÈ.(2)Ö¿½Âf(0)=0.π1duf(x)3[0,1]þk.§qsgnsin3x=0,(n=1,2,···)mä§d K£1¤§f(x)3[0,1]þxnÈ.∗2.e¼êf(x)3[a,b]þȧÙÈ©´I§83[a,b]Sk:þUCf(x)¦§¤,¼êf(x)§y∗²f(x)3[a,b]þȧ¿ÙÈ©EI.∗y²µ-F(x)=f(x)−f(x)§KF(x)3[a,b]þØUCf(x)¼êk:þ0§=Øùk:§¼êëY§lF(x)3[a,b]þȧȩ0.∗qf(x)=f(x)−F(x)§âȼêEȧZZZZbbbb∗kf(x)dx=f(x)dx−F(x)dx=f(x)dx=I.aaaa23.?Øf,f,|f|nömÈ5"X.)µ课后答案网2(1)ef(x)3[a,b]þȧKf(x)3[a,b]þÈ.Ïf(x)3[a,b]þȧf(x)3[a,b]þ7k.f(x)6M,M~ê(x∈[a,b])000200200000003«m[xi−1,xi]þ?ü:x,x§Äf(x)−f(x)=[f(x)−f(x)][f(x)+f(x)]20020ωiL«f(x)3[xi−1,xi]þÌݧK|f(x)−f(x)|62ωiMXX2ef(x)3[xi−1,xi]þÌÝΩi§ÒkΩi62ωiM§lkΩi∆xi62Mωi∆xiXXiiwww.hackshp.cn2duf(x)ȧkωi∆xi→0(λ(∆)→0)§KΩi∆xi→0(λ(∆)→0)§ùÒ`²f(x)3[a,b]þiiÈ5.(2)ef(x)3[a,b]þȧK|f(x)|3[a,b]þÈ.∗©Or¼êf(x)|f(x)|3«m[xi−1,xi]þÌÝPωi,ωi.XÏéáu[x000000000∗∗i−1,xi]?¿ü:x,x§k||f(x)|−|f(x)||6|f(x)−f(x)|§kωi6ωi§u´ωi∆xi6Xiωi∆xiiXX∗duωi∆xi→0(λ(∆)→0)§Ò±íλ(∆)→0§kωi∆xi→0§ùÒ`²|f(x)3[a,b]þiiÈ5.(3)e|f(x)|3[a,b]þȧØU½f(x)3[a,b]þÈ.1,xknê~µf(x)=−1,xÃnê|f(x)|=13?Û4«mþȧf(x)%3?Û4«mþÑØÈ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn1192(4)ef(x)3[a,b]þȧØU½f(x)3[a,b]þÈ.1,xknê~µf(x)=−1,xÃnê2f(x)=13?Û4«mþȧf(x)%3?Û4«mþÑØÈ.22(5)e|f(x)|3[a,b]þȧKd(1)§f(x)=|f(x)|3[a,b]þ½È.p(6)ef2(x)3[a,b]þȧKd|f(x)|=f2(x)§ ¼êëY§p ¼êȧd¼ê7È.ZZbb4.e¼êf(x)3[a,b]þȧy²3ò¼êϕn(x)(n=1,2,3,···)¦f(x)dx=limϕn(x)dxn→∞aa(n)(n)(n)(n)(n)iy²µò[a,b]n ©§©:a=x00§3[A,B]þëY¼êϕ(x)§¦|f(x)−ϕ(x)|dx0§3δ>0§é[A,B]¥?¿ü:x,x§|x−00000εx|<δ§Òk|ϕ(x)−ϕ(x)|<2(b−a)ZZb课后答案网bu´|h|<δ§k|f(x+h)−f(x)|dx=|f(x+h)−ϕ(x+h)+ϕ(x+h)−ϕ(x)+ϕ(x)−f(x)|dx6ZaZaZZbbbB|f(x+h)−ϕ(x+h)|dx+|ϕ(x+h)−ϕ(x)|dx+|f(x)−ϕ(x)|dx6|f(x+h)−ϕ(x+h)|dx+ZaZaZaAbBbεεε|ϕ(x+h)−ϕ(x)|dx+|f(x)−ϕ(x)|dx<+dx+=εaZA4a2(b−a)4bwww.hackshp.cnllim|f(x+h)−f(x)|dx=0h→0a若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn120§3½È©5ZZZbbb1.ef(x),g(x)3[a,b]ȧy²f(x)+g(x)3[a,b]ȧ¿(f(x)+g(x))dx=f(x)dx+g(x)dx.ZZaaabby²µÏf(x),g(x)3[a,b]ȧ=f(x)dx,g(x)dx3§é?¿©{∆:a=x00§f(x)Øð"§y²f(x)dx>0.ay²µÏf(x)>0Øð"§K73x0∈[a,b]§¦f(x0)>0x0−ab−x0f(x0)dëY¼êÛÜÒ5§30<δ6min,§¦x∈[x0−δ,x0+δ]§f(x)>>222ZZZbx0+δx0+δf(x0)0§u´kf(x)dx>f(x)dx>dx=f(x0)δ>0.ax0−δ课后答案网x0−δ24."eK¥È©µZZ112(1)xdx,xdx00ZπZπ22(2)xdx,sinxdx00Z−1xZ1www.hackshp.cn1x(3)dx,3dx−230)µZZ1122(1)Ïx∈(0,1)§x>x§Kxdx>xdx00ZπZππ22(2)Ïx∈0,§x>sinx§Kxdx>sinxdx200Z−1xZ1x−2Z1112−x2−xx(3)Ïdx=dx=3dxx∈(0,1)§2−x>x§3>3§−Z230Z30−1x11xldx>3dx−230Zb25.f(x)3[a,b]ëY§f(x)dx=0§y²f(x)3[a,b]þð".a2y²µ^y{.bf(x)3[a,b]þØð"§Kf(x)>0Øð".若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn1212qf(x)3[a,b]ëY§f(x)3[a,b]ëY§ZZbb22Kâ13Kf(x)dx>0§ù®f(x)dx=0gñ.aau´bاlf(x)3[a,b]þð".26.Þ~`²µf(x)3[a,b]ȧf(x)3[a,b]ØÈ.1,xknê2)µ~µf(x)=§f(x)=1§−1,xÃnê22Kf(x)3[a,b]þëY§lf(x)3[a,b]þÈqd1K§¼êf(x)3[a,b]þØÈ.XnZb7.f(x),g(x)3[a,b]ëY§y²limf(ξi)g(θi)∆xi=f(x)g(x)dx§Ù¥xi−16ξi6xi,xi−16max(∆xi)→0ai=1θi6xi(i=1,2,···,n),∆xi=xi−xi−1(x0=a,xn=b).Xny²µÏf(x),g(x)3[a,b]ëY§Kf(x)·g(x)3[a,b]ëY§f(x)·g(x)3[a,b]ȧ=limf(ξi)g(ξi)∆xi=max(∆xi)→0i=1Zbf(x)g(x)dxaXnXnlimf(ξi)g(θi)∆xi=limf(ξi)g(ξi)∆xi+max(∆xi)→0max(∆xi)→0"i=1i=1#XnXnlimf(ξi)g(θi)∆xi−f(ξi)g(ξi)∆ximax(∆xi)→0i=1i=1XnXnXnXnqf(ξi)g(θi)∆xi−f(ξi)g(ξi)∆xi=f(ξi)(g(θi)−g(ξi))∆xi6|f(ξi)||g(θi)−g(ξi)|∆xi6i=1i=1i=1i=1XnXnM(f)ωi(g)∆xi=M(f)ωi(g)∆xi§Ù¥M(f)L«|f|3[a,b]þþ.§ωi(g)L«g3[xi−1,xi]þi=1i=1Ì.XnXndfëY5ÚgÈ5§max(∆xi)→0§þ¡Ø ªmàM(f)ωi(g)∆xi→0§llimf(ξi)g(θi)∆xi=max(∆xi)→0i=1i=1Zbf(x)g(x)dx.aZa8.y=ϕ(x)(x>0)´îüNOëY¼ê§ϕ(0)=0,x=ψ(x)´§¼ê§y²ϕ(x)dx+Z0bψ(y)dy>ab(a>0,b>0).课后答案网ay²µdy=ϕ(x)´îüNOëY¼ê§ϕ(0)=0Ù¼êx=ψ(y)´îüNOëY¼ZZabê§ψ(0)=0§Ïϕ(x)dx,ψ(y)dyk½ÂZ00x-g(x)=bx−ϕ(t)dt§AO/§k0Zwww.hackshp.cnag(a)=ab−ϕ(t)dt(1)00g(x)=b−ϕ(x)00dϕ(x)´îüNOëY¼ê§Ïd00¶x>ψ(b)§kg(x)<0¶0x=ψ(b)§kg(x)=0§Ïdx=ψ(b)§g(x)§=kg(a)6maxg(x)=g(ψ(b))(2)ZZψ(b)ψ(b)0©ÜÈ©§xϕ(x)dx=bψ(b)−ϕ(x)dx=g(ψ(b))§^Cþy=ϕ(x)§Kx=ψ(y)§u´00ZZψ(b)b0g(ψ(b))=xϕ(x)dx=ψ(y)dy(3)00ZZabò(??)!(??)(??)Òϕ(x)dx+ψ(y)dy>ab(a>0,b>0).0a若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn122§4½È©O1.Oe½È©µZ22(x+1)(x−3)(1)dx3x21Zπ2(2)(asinx+bcosx)dx1Z14x−1(3)dx0x+1Z12x+1(4)dxx4+10Z√153210(5)x(1−5x)dx0Z1222(6)x(2−3x)dx0Z15√(7)x2−5xdx−15Zπ2(8)sinmxcosnxdx0Z1318x−4(9)√dx−19x2+6x+53Z√ln23−x2(10)xedx0Z1(11)xarctanxdx0Z2π2(12)xcosxdx0Zπ2(13)xcosxdx课后答案网−πZ2πω(14)sinωtsin(ωt+ϕ)dt0Z3xdx(15)√01+1+xZ4√www.hackshp.cn(16)x(x+x)dx0Zπ(17)sinmxsinnxdx−πZπ(18)sinmxcosnxdx−πZ0p(19)(x+1)1−x−x2dx−1Z0.75dx(20)√0(x+1)x2+1)µZZ22232(x+1)(x−3)x+x−3x+3(1)dx=dx=13x213x2Z2221331x313x+1−x−x2dx=32+x−3lnx+x=3−ln211若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn123Zπ2π(2)(asinx+bcosx)dx=(−acosx+bsinx)|2=a+b01Z14Z24Z2x−1y−28243216(3)-y=x+1§Kdx=dy=1−+−+dy=0x+11y1yy2y3y4224161617y−8lny−+−=−8ln2yy23y331ZZ121x+1111(4)dx=√+√dx=0x4+120x2+2x+1x2−2x+1√√1√√√√12√√2[2arctan(2x+1)+2arctan(2x−1)]=(arctan(2+1)+arctan(2−1))=π202411(5)-x=√sinu,dx=√cosudu§55Z√1ZπZππ512121cos22ucos24u2Kx3(1−5x2)10dx=sin3ucos21udu=−(1−cos2u)cos21udcosu=−−=0250250252224016600ZZ1122224623(6)x(2−3x)dx=(4x−12x+9x)dx=00105√(7)-u=2−5xZ1Z√5√232√23Kx2−5xdx=(2u−u)du=(33−7)−12513755ZπZπcosm+nπ212m2(8)m6=±n§sinmxcosnxdx=[sin(m+n)x+sin(m−n)x]dx=−−020m2−n22(m+n)m−ncosπ2¶2(m−n)ZπZππ212121m=±nm6=0§sinmxcosnxdx=±sin2nxdx=−cos2nx=±(1−0204n04n1ncosnπ)=±[1−(−1)](n∈Zn6=0)¶4Znπ2m=0§sinmxcosnxd=00Z1Z1Z12318x−43d(9x+6x+5)103dx(9)√2dx=课后答案网√2−q =−19x+6x+5−19x+6x+53−13x+123331+2s123p103x+13x+1√10√29x2+6x+5−ln+1+=4(2−1)−ln(2+1)3223−132(10)-u=xZ√Zwww.hackshp.cnZln221ln21ln2ln213−x−u−u−uxedx=uedu=−ue+edu=(1−ln2)0202004Z121Z121(11)xarctanxdx=xarctanx−1xdx=π−1+1arctanx=π−202201+x2822400Z2πZ2π22π2πZ2π(12)xcos2xdx=1x(1+cos2x)dx=x+1xsin2x−1sin2xdx=π2020444000ZπZππZππZπ222(13)xcosxdx=2xcosxdx=2xsinx−4xsinxdx=4xcosx−4cosxdx=−4π−π00000Z2πZ2π2πω1ωπ1ωπ(14)sinωtsin(ωt+ϕ)dt=[cosϕ−cos(2ωt+ϕ)]dt=cosϕ−sin(2ωt+ϕ)=cosϕ020ω4ω0ωZ3Z3√Z3√3xdxx(1−1+x)235(15)√=dx=(1+x−1)dx=(1+x)2−3=01+1+x0−x0303ZZ4√435122(16)x(x+x)dx=(x+x2)dx=0015若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn124ZπZππ1sin(m−n)x(17)m6=±n(m,n∈Z)§sinmxsinnxdx=[cos(m−n)x−cos(m+n)x]dx=−−π2−π2(m−n)−πsin(m+n)xπ=0¶2(m+n)−πZZZπππ2m=±n(m,n∈Z)m6=0§sinmxsinnxdx=±sinmxdx=±(1−cos2mx)dx=−π−π0±π¶Zπm=0§sinmxsinnxdx=0−πZπ(18)sinmxcosnxdx=0−πZZZZ0p00p0p212121212(19)(x+1)1−x−xdx=−(1−x−x)2d(1−x−x)+1−x−xdx=1−x−xdx=−1√2−12−12−1155+arcsin485(20)-x=tantZZ0.75arctan0.75dxdtK√==0(x+1)x2+10sint+costπ√arctan0.75π√Zarctan0.75dt+arctan0.75tan+4=2lntant+π=√1ln28=√1ln9+420√π2282tanπ272sint+0482.OeÈ©µZπ27(1)sinxdx0Zπ24(2)cosxdx0Zπ5(3)sinxdx0Z2π6(4)cosxdx0Za22n(5)(a−x)dx0课后答案网Z126(6)(1−x)dx0)µZπ276!!(1)sinxdx=07!!Zπwww.hackshp.cn243!!π3π(2)cosxdx=·=04!!216ZZπZππ5255(3)sinxdx=sinxdx+sinxdx00π2ZZπ0553 È©¥§-x=π−y§Ksinx=siny,dx=−dy§u´sinxdx=−sinydy=ππZ22π25sinydy=I50Zπ54!!16lsinxdx=2I5=2·=05!!15ZZZZπ2πππ666265!!π15(4)cosxdx=cosxdx=2cosxdx=4cosxdx=4··=π0−π006!!224(5)-x=acost,dx=−asintdt§ZZZπa022n2n+12n+12n+122n+1(2n)!!2n+1K(a−x)dx=−asintdt=asintdt=a0π0(2n+1)!!2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn125Z12612!!(6)3þK¥§-a=1,n=6§K(1−x)dx=013!!ZZa+nTT3.f(x)´±Ï¼ê§±Ï´T§y²f(x)dx=nf(x)dx§d?n´ê.ZZaZ0ZZa+nT0TnTa+nTy²µf(x)dx=f(x)dx+f(x)dx+···+f(x)dx+f(x)dxaa0(n−1)TnTZZZa+nTaaéþã ª È©§x−nT=t§Kf(x)dx=f(t+nT)dt=f(t)dtZnTZ0Z0iTiTTé10)020ZaZa2Za223211y²µ-t=x§K2xdx=dt§u´xf(x)dx=tf(t)dt=xf(x)dx02020ZZZxxu9.|^©ÜÈ©y²µf(u)(x−u)du=f(x)dxduZZ0Z0x0ZZZZxuuxxxxy²µf(x)dxdu=uf(x)dx−uf(u)du=xf(t)dt−uf(u)du=xf(u)du−Z00Z000000xxuf(u)du=f(u)(x−u)du00210.Ýlîù§¤É1ÖU5Æp(x)=a+bx+cx©Ù§Ádeã^û½Xêa,b,c¶o1Ö´P=Zl2p(x)dx§41Öul?§34:mü>«Éo1Ö.03)µd®§Zlb2c3(1)P=p(x)dx=al+l+l023Zx022242(2)-P(x)=p(t)dt§KPl=pl=a+bl+cl=003339Z2l32283P(3)p(x)dx=al+bl+cl=039812b2c3al+l+l=P23课后答案网2283Péá§|al+bl+cl=39812242a+bl+cl=0394a=Pl69¦)§b=−4l2Pwww.hackshp.cn135c=P8l311.ef(x)ëY§¦µZbd(1)f(t)dtdxx!Zx2d(2)f(t)dtdx0)µZZZbxxddd(1)f(t)dt=−f(t)dt=−f(t)dt=−f(x)dxxdxbdxb!!!Zx2Zx2Zx22d2dddx2(2)Ïf(t)dt=f(x)§f(t)dt=f(t)dt·=2xf(x)dx20dx0dx20dx12.¦4µ若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn12712n−1(1)lim++···+n→∞n2n2n2ppp1+2+···+n(2)lim(p>0)n→∞np+1rrr!112n−1(3)lim1++1++···+1+n→∞nnnn√nn!(4)limn→∞n)µi1(1)¼êf(x)=x3[0,1]ëY§ÏÈ.ò[0,1]n ©§©:,∆xi=(i=0,1,···,n−1)§3znnii+1ii«m[xi,xi+1]=,§ξi=§Kf(ξi)=§nnnnnX−1Z112n−1112n−1i1u´lim++···+=lim++···+=lim·=xdx=n→∞n2n2n2n→∞nnnnn→∞nn0i=012pi1(2)¼êf(x)=x3[0,1]ëY§ÏÈ.ò[0,1]n ©§©:,∆xi=(i=1,2,···,n)§3z«pnni−1iiim[xi−1,xi]=,§ξi=§Kf(ξi)=§nnnn1p+2p+···+npXnip1Z11pu´lim=lim·=xdx=n→∞np+1n→∞nn0p+1i=1√i1(3)¼êf(x)=1+x3[0,1]ëY§ÏÈ.ò[0,1]n ©§©:,∆xi=(i=0,1,···,n−1)§3rnnii+1iiz«m[xi,xi+1]=,§ξi=§Kf(ξi)=1+§nnnnrrr!nX−1rZ1112n−1i11√u´lim1++1++···+1+=lim1+·−lim=x+1dx−n→∞nnnnn→∞nnn→∞n0i=02√0=(22−1)3√r11√n1nn!nn!1n2nnnn!112n(4)Ï==···§ln=ln+ln+···+lnnnnnnnnnnnnZ1课后答案网i1q¼êf(x)=lnx3(0,1]ëY§Älimlnxdx.ò(0,1]n ©§©:,∆xi=(i=1,2,···,n)§ξ→+0ξnnpi−1iii3z«m[xi−1,xi]=,§ξi=§Kf(ξi)=§nnnn√nXnZ1n!112ni1u´limln=limln+ln+···+ln=limln·=limlnxdx=n→∞nn→∞nnnnn→∞nnξ→+0ξi=11lim(xlnx−x)|ξ=−1www.hackshp.cnξ→+0√nn!−11llim=e=n→∞neI2nπ22442n2n13.â~7klim=1§díy=··········n→∞I2n+1213352n−12n+1ZπZπ22n(2n−1)(2n−3)···3·1π22n+12n(2n−2)···4·2y²µÏI2n=sinxdx=·,I2n+1=sinxdx=§02n(2n−2)···4·220(2n+1)(2n−1)···5·3I2n+122442n2n2K=········I2n13352n−12n+1πZπZππ22n+122n2n+12n2n−106x6§06sinx61,sinx6sinx6sinx§Ksinxdx6sinxdx6Z200π22n−1I2nI2n−1sinxdx=I2n+16I2n6I2n+1§u´1660I2n+1I2n+1n−12nI2n+12nI2n+12nqd4íúªIn=In−2,I2n+1=I2n−1==§lim=lim=n2n+1I2n−12n+1n→∞I2n−1n→∞2n+1I2nπ22442n2n1§u´lim=1§l=··········n→∞I2n+1213352n−12n+1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn12814.f(x)g(x)Ñ3[a,b]ȧy²Z2ZZbbb22f(x)g(x)dx6f(x)dx·g(x)dxaaaq ª3Û¤áº2222y²µé?Û¢êh§Ï[hf(x)−g(x)]=hf(x)−2hf(x)g(x)+g(x)>0ZZZZbbbb222222dÈ©5§(hf(x)−2hf(x)g(x)+g(x))dx>0=hf(x)dx−2hf(x)g(x)dx+g(x)dx>aaaa0Z2ZZZ2bbbb22dgnªK^§2f(x)g(x)dx−4f(x)dx·g(x)dx60=f(x)g(x)dx6ZZaaaabb22f(x)dx·g(x)dxaaZ2ZZZZbbbbb2222¦ Ò¤á§2f(x)g(x)dx−4f(x)dx·g(x)dx=0=hf(x)dx−2hf(x)g(x)dx+Zaaaaab2g(x)dx=0k.aZZZZbbbb2222Øh0§§Kh0f(x)dx−2h0f(x)g(x)dx+g(x)dx=0=[h0f(x)−g(x)]dx=aaaa0ZZZbbb222g(x)=h0f(x)§[h0f(x)−g(x)]dx=0(Ù¥h0§hf(x)dx−2hf(x)g(x)dx+Zaaab2g(x)dx=0)a课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn1291lٽȩA^ÚCqO§1²¡ã/¡È1.¦de¤¤ã/¡Èµ22(1)y=4(x+1),y=4(1−x)(2)y=|lnx|,y=0,(0.16x610)2(3)y=x,y=x+sinx,(06x6π)2(4)y=1−x,2y=x+2(5)]r=acosθ+b(b>a)§b=a=%9(6)r=3cosθ,r=1+cosθ(7)^Óx=a(t−sint),y=a(1−cost)(06t62π)±9x¶222(8)(/x3+y3=a3)µ222y−4−y+4y−4(1)ü^x=x==−:pI©O−292§Z44Z42222y−4y−4216u´A=−−dy=−(y−4)dy=−24403(2)ü^y=|lnx|y=0:îI1§Z10Z1Z10110u´A=[ln|x|−0]dx=(−lnx)dx+lnxdx=−(xlnx−x)+(xlnx−x)=0.10.110.119.9ln10−8.1≈14.69559ZπZπZππ(3)A=(x+sin2x−x)dx=sin2xdx=1(1−cos2x)dx=π−sin2x=π00202420(4)ü^:©O(0,1),(−8,−3)§ZZ112232u´A=[1−y−(2y−2)]dy=(3−y−2y)dy=−3−33ZZ2π2π1212π22(5)¤¦¡ÈµA=rdθ=(acosθ+b)dθ=a+πb2020223课后答案网9(6)¤¦¡ÈµA=π−A1=π−A124ZπZπ132232Ù¥A1=[9cosθ−(1+cosθ)]dθ=[8cosθ−1−2cosθ)dθ=π§2−π035lA=π4ZZZ2π2π2π222(7)¤¦¡ÈµA=ydx=www.hackshp.cna(1−cost)·a(1−cost)dt=a(1−cost)dt=3πa00033π(8)x=acost,y=asint§Ù¥06t6§éAuo©¡È§¤¦¡ÈÙoZZ3Zπa024222463π2=A=4ydx=4(−3asintcost)dt=12a(sinx−sinx)dx=a0π08222A2.y=xrýx+3y=6y¡È©¤üÜ©A(¬)ÚB(¬)§¦.Bx2√2)µd®§ý§+(y−1)=1§Ký¡ÈS=πab=3π3223qy=xýx+3y=6y:pI0,2Z3Z33√2p√2py2233u´A=(6y−3y2−y)dy=31−(y−1)2dy−=π−§KB=S−A=0020342√33π+§34√A43π−9l=√.B83π+9若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn130√3.¦y=√1−x2+arccosxx¶9x=−1¤¡È.)µÏy1=1−x2½Â[−1,1]§[0,1]¶y2=arccosx½Â[−1,1]§[0,π]ZZZZ111p13K¡ÈA=y1dx+y2dx=1−x2dx+arccosxdx=π−1−1−1−12课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn131§2l¦elµ31.y=x2(06x64)1212.x=y−lny(16y6e)422223.(/x3+y3=a3(a>0)4.^Óx=a(t−sint),y=a(1−cost)(06t62π)5.ìmx=a(cost+tsint),y=a(sint−tcost)(06t62π)6.%9r=a(1+cosθ)(06θ62π))µZqZr44√3981.¤¦ls=1+[(x2)0]2dx=1+xdx=(1010−1)00427ssZepZe2Ze2Zey1y1y12.¤¦ls=1+(x0)2dy=1+−dy=+dy=+dy=1122y122y122y2e+14333.d®x=acost,y=asint(06t62π)Zπq2K¤¦ls=4[(acos3t)0]2+[(asin3t)0]2dt=Z0πqZπ224a(−3cos2tsint)2+(3sin2tcost)2dt=12asintcostdt=6a00Z2πp4.s=[(a(t−sint))0]2+[(a(1−cost))0]2dt=Z0qZrZ2π2π2π221−costt|a|(1−cost)+sintdt=2|a|dt=2|a|sindt=8|a|00202Z2πp5.s=[(a(cost+tsint))0]2+[(a(sint−tcost))0]2dt=Z0Z2πp2π|a|(tcost)2+(tsint)2dt=|a|tdt=2π2|a|00Z2πpZ2πq课后答案网ZπrZπ6.s=r2+(r0)2dθ=a2(1+cosθ)2+a2sin2θdθ=4|a|1+cosθdθ=4|a|cosθdθ=0002028|a|www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn132§3NÈ1.¦Ñde¡¤¤AÛNNȵ(1)¦ INNȧÙþe.ý§ý¶©O uA,BÚa,b§ph¶222xyz(2)¦ý¥NNȵ++=1¶a2b2c2222222(3)¦deü¡µx+y+z=a,x+y=ax¤¤Nȶ22(4)¦^ÏL.¡»²¡lÎþeü/NNȧÎ.»a§.¡§x+y62a§ ¡ÏLx¶þ».¡¤α.)µ0x(1)²1uþ!e.åle.x ¡§d ¡ý§Ù¶©Oµa=A+1−(a−A),h0xb=B+1−(b−B)h00u´d ¡¡ÈµA(x)=πab=2xxπAB+(a−A)(b−B)1−+(A(b−B)+B(a−A))1−hhZhπl¤¦NÈV=A(x)dx=[(2a+A)b+(a+2A)B]0622yz(2)^RuOx¶²¡ ý¥ èý§3yOz²¡þÝK+=22xxb21−c21−a2a21rr222xxxdÙ¶©Ob1−9c1−§ldý¡ÈA(x)=πbc1−a2a2a2u´§¤¦ý¥NNȵZZaa2x4V=A(x)dx=2πbc1−dx=abca23−a0p√√(3)z=a2−x2−y2(þ¡)§ÙCz−ax−x26y6ax−x2√Zax−x2pr222311322xÙ ¡ÈA(x)=2a−x−ydy=a2x2−a2x2+(a−x)arcsin0a+xu´§¤¦NȵZa课后答案网V=2A(x)dx=Z0ra311322x2342a2x2−a2x2+(a−x)arcsindx=aπ−0a+x33√√(4)y=a2−x2,z=a2−x2tanα§1p22p22122KA(x)=a−x·a−xtanα=(a−x)tanα2www.hackshp.cn2l¤¦ZNȵZaa2223V=A(x)dx=(a−x)tanαdx=atanα−a032.¦^=NNȵ22xy(1)+=17X¶a2b2(2)y=sinx,y=0(06x6π)(i)7x¶(ii)7y¶33(3)x=asint,y=bcost(06t62π)(i)7x¶(ii)7y¶(4)y²da6x6b,06y6y(x)(Ù¥y(x)´ëY¼ê)¤¤¡È7y¶^=¤^=NNȵZbV=2πxy(x)dxa若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn133(5)x=a(t−sint),y=a(1−cost)(06t62π,y=0)(i)7x¶(ii)7y¶(iii)7y=2a)µZZaa222x42(1)V=πydx=πb1−dx=πaba23−a−aZπ22π(2)(i)V=πsinxdx=02Zπ2(ii)V=2πxsinxdx=2π0ZπZπZπZπ2222622227227(3)(i)V=2πydx=2πbcost·3asintcostdt=6aabsintcostdt=6πab(cost−00009322cost)dt=πab105(ii)|^é¡5§IòþªY¥a,béN§=322V=πab105(4)y²µ[a,b]?¿©{µa=x0÷/74¶^=πr(θi)sinθi∆θi333i=1XnZβ23002π3lV=limπr(θi)sinθi∆θi=r(θ)sinθdθ.λ→033αi=1Zπ23383(2))µV=πa(1+cosθ)sinθdθ=πa3034.rÔy=x(x−a)3îI0c(c>a>0)mlã7x¶^=§¯cÛ§T^=NNÈV u±uOP7x¶^=¤INNȺ(ã8-14))µÏÔy=x(x−a),xP=c§P(c,c(c−a))12π32K±uOP7x¶^=¤INNȵV1=πc[c(c−a)]=c(c−a)33若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn134Zc522ca4a3¤¦^=NNȵV2=π[x(x−a)]dx=π−c+c052352π32ca4a3qV1=V2§c(c−a)=π−c+c§35235lc=a4√x5.ry=7x¶^=^=N§§3:x=0x=ξmNÈPV(ξ)§¦a uÛ§U1+x21¦V(a)=limV(ξ).2ξ→∞Zξ√222xξa)µÏV(ξ)=πdx=π§KV(a)=π01+x22(1+ξ2)2(1+a2)211ξπ2qV(a)=limV(ξ)=limπ=§u´a=12ξ→∞2ξ→∞2(1+ξ2)4qa>0§a=1.2222226.ýbx+ay=ab7x¶^=^=ý¥N§r§÷x¶B%§¦e/NNÈ uý¥NNȧû½}»ρ(ã8-15).)µK¥e/NNÈV√§ý¥NNÈV1222222a2b2−b2x2Ïbx+ay=ab§Ky=aZZaa22222ab−bx42KV1=πydx=πdx=πaba23−ap−aa2b2−a2ρ2Za√a2b2−b2x22V=V2−2π√1−2πρdx=ba2b2−a2ρ2ab4p224πa2p224a223πabb−ρ−ρb−ρ=π(b−ρ)233b3bq14a22322−2dK¿§V=V1=π(b−ρ)2=πab§)d§ρ=b1−2323b3课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn135§4^=¡¡È1.¦e^=¡¡Èµ2(1)x=2py+a(06x6a,a>1)7x¶9y¶(2)y=sinx(06x6π)7x¶22xy(3)ý+=17y¶a2b2(4)^Óx=a(t−sint),y=a(1−cost)(06t62π)7x¶22(5)VÝr=2acos2θ(i)74π(ii)7¶θ=2π(iii)7¶θ=4)µ22x−aaa−a(1)y=,−6y6(p>0)2p2p2pvZa2uu"20#2Zap(i)Fx−atx−aπ2222x=2π1+dx=(x−a)x+pdx=02p2pp20"p#a(2a2−4a+p2)pp2+4aa+a2+p2p2+a2−lnπ8p28pppp2+2py+a(ii)1+(x0)2=√2py+aZa2−ap2ppp2+2py+aKFy=2π2py+a√dy=−a2py+a2p2Za−aπ2pp2222π223222py+p+ad(2py+p+a)=(p+a)2−pπp−a3p32pZZπpπp√√(2)F=2πsinx·1+[(sinx)0]2dx=2πsinx1+cos2xdx=22π+2πln(2+1)00vZZu!2Zrbpbapua(−y)aba2−b2(3)F=2πx1+(x0)2dy=2πb2−y2·t1+pdy=2πb2+y2dy=−b课后答案网−bbbb2−y2b−bb2Zrb224aπc2ba+cb2+ydy=2aπa+ln.b0b2cbZZpt2π2πt(4)ÏdS=(x0)2+(y0)2dt=2asindt§KF=2πyds=2πa(1−cost)·2asindt=Z20022π23tt64216aπsind=πawww.hackshp.cn.0223√√√2aππ(5)(i)y=2acos2θsinθ,dS=√dθ−6θ6cos2θ44Zπ4√22dé¡5§F=2×2π2asinθdθ=4πa(2−2)0√√√2aππ(ii)x=2acos2θcosθ,dS=√dθ−6θ6cos2θ44Zπ4√22KF=2π2acosθdθ=42πa−π√√4√√(iii)x=2acos2θcosθ,y=2acos2θsinθ,√2aππdS=√dθ−6θ6cos2θ44ππ5¿3−6θ6S§ðkx−y>0§44Zπ√Zπ4x−y2a√242u´§¤¦L¡ÈF=2×2π√·√dθ=42πa(cosθ−sinθ)dθ=8πa−π2cos2θ−π44若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn136ZTp2.y²dx=ϕ(t),y=ψ(t),z=χ(t)(t06t6T)Oxy²¡m¤Î¡¡È uS=χ(t)ϕ02(t)+ψ02(t)dtt0x=ϕ(t)y²µCD3Oxy²¡ÝKAB§KAB§(t06t6T)y=ψ(t)3ABþ©:µA=M0,M1,···,Mi−1,Mi,···,Mn=B3CDéA©:µC=N0,N1,···,Ni−1,Ni,···,Nn=DéAëêµt0,t1,···,ti−1,ti,···,tnMiIχi=ϕ(ti),yi=ψ(ti)§KMiNi=χ(ti)F/MiMi−1Ni−1Ni¡ÈµMiNi+Mi−1Ni−1χ(ti)+χ(ti−1)pSi=·Mi−1Mi=[ϕ(ti)−ϕ(ti−1)]2+[ψ(ti)−ψ(ti−1)]2=p22χ(ti)[ϕ(ti)−ϕ(ti−1)]2+[ψ(ti)−ψ(ti−1)]2−χ(ti)−χ(ti−1)p[ϕ(ti)−ϕ(ti−1)]2+[ψ(ti)−ψ(ti−1)]22000d©¥½nµϕ(ti)−ϕ(ti−1)=ϕ(ξi)∆ti,ψ(ti)−ψ(ti−1)=ψ(ηi)∆ti,χ(ti)−χ(ti−1)=χ(ζi)∆tip1q§S0202002022i=χ(ti)ϕ(ξi)+ψi(ηi)∆ti−2χ(ζi)ϕ(ξi)+ψi(ηi)(∆ti)lΡ¡ÈµXnq1Xnqϕ02(ξ02002022S=limχ(ti)i)+ψi(ηi)∆ti−limχ(ζi)ϕ(ξi)+ψi(ηi)(∆ti)λ→0λ→02i=1i=1ZZTpTp=χ(t)ϕ02(t)+ψ02(t)dt−0=χ(t)ϕ02(t)+ψ02(t)dtt0t0课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn137§5%1.¦eã%Iµ1(1)»a§laα(α6π)þ! l¶2(2)±A(0,0),B(0,1),C(2,1),D(2,0)º:Ý/±.§þ?:Ý uT::ål¶kθ(3)éêÚr=ae(a>0,k>0)þd:(0,a):(θ,r)þ!lã.)µ(1)±:%§l²å©>¤3x¶ïáIX§K l§x=acosα,y=00asinα§u´x=−asinα,y=acosα§ZαZα2p2acosαx02+y02dαa2cosαdαa2sinα2asinαlx=0=0=2=2ss1αaαZαZα22p2asinαx02+y02dαa2sinαdα002aαy===1−cosssα2(2)k¦Ñݼê.x=0AB§y∈[0,1]§Ùþ?:Ýy=yρ=2yx=xBC§x∈[0,2]§Ùþ?:Ýy=1√ρ=21+x2x=2CD§y∈[0,1]§Ùþ?:Ýy=ypρ=24+y2x=xDA§x∈[0,2]§Ùþ?:Ýy=0ρ=2xZ1pu´m=2y02+12dy=1,AB0Z课后答案网1py·2y02+12dy02x=0,y==¶ABABm3ZAB2pp√√m=21+x212+02dx=25+ln(2+5),BCZ02ppx·21+x212www.hackshp.cn+02dx√02(55−1)x==√√,BCmBC65+3ln(2+5)y=1¶BCZ√1pp√1+5m=24+y202+12dy=5+4ln,CD20Z1ppy·24+y202+12dy√02(55−8)xCD=2,y==√¶CDm√1+5CD35+4ln2Z2pm=2x12+02dx=4,DAZ02px·2x12+02dx04x==,y=0DAm3DADAdÝ/±.%Iµmx+mx+mx+mxx=ABABBCBCCDCDDADA=m+m+m+mABBCCDDA若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn138√√1+5165+14+24ln2√√√1+595+15+3ln(2+5)+12ln2my+my+my+myy=ABABBCBCCDCDDADA=m+m+m+m√ABBC√CDDA165−14+3ln(2+5)√√√1+595+15+3ln(2+5)+12ln2(3)%IµZZθpθpxx02+y02dxrcosθ[(rcosθ)0]2+[(rsinθ)0]2dθx=0=0=Zsθp[(rcosθ)0]2+[(rsinθ)0]2dθZ0Zθpθrcosθa2(1+k2)ekθdθae2kθcosθdθ2kθ200ake(sinθ+2kcosθ)−2akZ=Z=θpθ(4k2+1)(ekθ−1)a2(1+k2)ekθdθekθdθ002kθake(2ksinθ−cosθ)+akÓ{y=(4k2+1)(ekθ−1)u´§%4Iµpakpr=x2+y2=(e4kθ+1−2e2kθcosθ)(4k2+1)(4k2+1)(ekθ−1)2kθye(sinθ+2kcosθ)−2ktanθ0==xe2kθ(2ksinθ−cosθ)+1%I(r,θ0)2.^Ýþ!7áj¤»al§3üà^Ó7ájþü^£ã8-19¤§¯bõ§U¦7ájMABN%Ð3 %Oº)µy6"AM-a0x&课后答案网BNZ3π27ájݵ§lþµm=µds=πaµπ2π3plµx=acosθ,y=asinθ6θ6π,ds=x02+y02dθ=adθ22Kl%IZwww.hackshp.cn3πZ3π222xµdsµacosθdθππ2ax=2=2=−;mπaµπZ3Z3ππ222yµdsµasinθdθππy=2=2=0mπaµbqü^%Iµx=,y=0§þµ2bµ22ab−·πaµ+·2bµ√u´d®§:X%Iµx=π2=0§lb=2aπaµ+2bµ3.¶10§Ý©Ùρ=ρ(x)=(6+0.3x)Z/§Ù¥xå¶à:ål§¦¶þ.ZZ1010)µm=ρ(x)dx=(6+0.3x)dx=75(Z)0024.®Ôãy=x(−16x61)§ãþ?:?ÝT:y¶ål¤"§x=1?Ý5§¦dãþ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn140§6²þ!õ1.®6>´¥>{Rüà>ØUm§ªÇω§OÑ3Rþ²þõÇ(©Å6Ú Å6ü«¹?Ø).ZTZπ22222ωωUm2Um)µÅ6§Ñ3Rþ²þõǵP1=P(t)dt=cosωtdt=T02π0R2RZZ2πT221ωωUm2Um Å6§Ñ3Rþ²þõǵP2=P(t)dt=cosωtdt=T02π0R2Rhihiπππ2.O6>Øu=Umcosωt30,Ú−,S²þ.hiωZπ2ω2ωπωω)µ30,S²þµu=Umcosωtdt=0¶ωπ0hiZπππω2ω23−,S²þµu=Umcosωtdt=Um.2ω2ωπ−ππ2ω3.¦e¼ê3½«mS²þµ(1)y=sinx,[0,π]x(2)y=xe,[0,1])µZπ12(1)y=sinxdx=π0πZ1x(2)y=xedx=104.r.¤Iå¤".®ú6åU¦1f§¯r.10fõõº)µd½n§5¡EåFþx¤"=FZ=kx.Z1010d^§k=1§ÏF=x§u´¤¦õW=Fdx=xdx=50(Z·f¤=5J005.?ïxxêke3.Î/3»20§Y 27§3pÑY¡3§rYĦ§OÑå¤õ.235)µÏ∆W=πr·∆x·x·10g=10gπx∆xZ3058KW=10gπxdx=4.37×课后答案网10π(J)326.,Y¥¹´F/§þ.6§e.2§p10§¦Y/÷¹¤Éå.Y"1ë/.x)µÏ∆F=2xyg∆x=2gx3−∆xZ510x6KF=2gx3−dx=1.63×10(N)053www.hackshp.cn7.ÔNU5Æx=ct(c>0)$ħxL«3mtSÔN£Äål§0{åݲ¤"§¦ÔNlx=0x=a{å¤õ.302)µÏx=ct(c>0)§v=x=3ct224q0{åݲ¤"§Kf=kv(k~ê)§u´f=9kct1a3qxlx=0x=a§tlt=0t=§cZ1Z1(a)3(a)3c2423c62727KW=9kct·3ctdt=27kctdt=kc3a30078.»r¥Y¥§Y¡§¥"1§yò¥lY¥Ñ§õõºp)µÏ∆W=1·π(r2−(x−r)2)2∆x(2r−x)=π(4r2x−4rx2+x3)∆xZ2r22344KW=π(4rx−4rx+x)dx=πr03若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn141§7½È©CqOZ1dx1.^Ô/úª¦Cq(n=3).1+x20)µn=3§O4꧵1112x0=0,y0=1.0000;x1=,4y1=3.8919;x2=,2y2=1.8000;x3=,4y3=3.2000;x4=,2y4=632351.3846;x5=,4y5=2.3607;x6=1,y6=0.50006|Z^Ô/úª§k1dx1≈[y0+2(y2+y4)+4(y1+y3+y5)+y6]=0.78541+x21802.¦,Ê.¡È.Ê.Xã8-24¤«§x¶´§é¡¶§OA2§10 ©§ÿêâXe£üµf¤µx020406080100120140160180200y08.511.011.510.510.08.06.54.52.50)µ|^Ô/úª§k200202A≈[0+0+4(8.5+11.5+10.0+6.5+2.5)+2(11.0+10.5+8.0+4.5)]=×224≈1493.3(cm)6×533.3°20à¡þ§ÿþà6î ¡¡È.XJlàWéWz2§ÿàY ÝXeL¤µx02468101214161820y(Y )0.40.81.42.02.42.11.91.61.30.80.4(Y üµ)¦dà6î ¡¡È£ã8-25¤)µ|^Ô/úª§k2022A≈[0.4+0.4+4(0.8+2.0+2.1+1.6+0.8)+2(1.4+2.4+1.9+1.3)]=×44≈29.3(m)6×53课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn1421n?êØ1Ü©ê?êÚ2ÂÈ©1ÊÙê?ê§1.ý£µþ4Úe41.y²µ(1)lim(xn+yn)6limxn+limynn→∞n→∞n→∞(2)lim(xn+yn)>limxn+limynn→∞n→∞n→∞y²µ(1)limxn,limynþkê§K{xn},{yn}þk.n→∞n→∞-αk=sup{xn},βk=sup{yn}"u´§n>k§kxn+yn6αk+βkn>kn>klsup{xn+yn}6αk+βkn>klim(xn+yn)=limsup{xn+yn}6limαk+limβk=limxn+limynn→∞k→∞n>kk→∞k→∞n→∞n→∞5µelimxn,limyn+∞.~Xµlimxn=+∞§¦"Xªm>{$k¿Â§KlimynØn→∞n→∞n→∞n→∞−∞.ù{xn}Ãþ.§{xn+yn}½Ãþ.§þã"Xªw,¤á¶elimxn=−∞§¦"Xªn→∞m>{$k¿Â§KlimynØ+∞.u´{yn}þk.§llim(xn+yn)=−∞§þã"Xªn→∞n→∞w,¤á.(2)Ïxn>inf{xn},yn>inf{yn}§xn+yn>inf{xn}+inf{yn}âe(.e.¥§Kinf{xn+yn}>inf{xn}+inf{yn}§linf{xn+yn}>inf{xn}+inf{yn}n>kn>kn>kKliminf{xn+yn}>liminf{xn}+inf{yn}=liminf{xn}+liminf{yn}k→∞n>kk→∞n>kn>kk→∞n>kk→∞n>k=lim(xn+yn)>limxn+limyn.n→∞n→∞n→∞2.xn>0,yn>0§y²µ课后答案网(1)limxnyn6limxn·limynn→∞n→∞n→∞(2)limxnyn>limxn·limynn→∞n→∞n→∞y²µ(1)Ï06xn6sup{xn},06yn6www.hackshp.cnsup{yn}§K06xnyn6sup{xn}·sup{yn}âþ(.´þ.¥§Kk06sup{xn·yn}6sup{xn}·sup{yn}l06sup{xn·yn}6sup{xn}·sup{yn}n>kn>kn>kKlimsup{xn·yn}6limsup{xn}·sup{yn}=limsup{xn}·limsup{yn}k→∞n>kk→∞n>kn>kk→∞n>kk→∞n>k=limxnyn6limxn·limyn.n→∞n→∞n→∞(2)Ïxn>inf{xn}>0,yn>inf{yn}>0§Kxnyn>inf{xn}·inf{yn}>0âe(.´e.¥§Kkinf{xn·yn}>inf{xn}·inf{yn}>0linf{xn·yn}>inf{xn}·inf{yn}>0n>kn>kn>kKliminf{xn·yn}>liminf{xn}·inf{yn}=liminf{xn}·liminf{yn}k→∞n>kk→∞n>kn>kk→∞n>kk→∞n>k=limxnyn>limxn·limynn→∞n→∞n→∞3.elimxn3§Ké?Ûê{yn}¤áµn→∞(1)lim(xn+yn)=limxn+limynn→∞n→∞n→∞若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn143(2)lim(xn·yn)=limxn·limyn§elimxn>0n→∞n→∞n→∞n→∞y²µlimxn=αn→∞elimyn=+∞(½−∞)§K(1)w,¤á.Ïα>0§K(2)w,¤á.n→∞Ølimyn=βkên→∞Ïlimyn=β§3{yn}f{yn}§¦limyn=ββ¤kÂñf4¥ö.kkn→∞k→∞qlimxn=α§Klimxn=α§lim(xn+yn)=α+β,lim(xn·yn)=αβkkkkkn→∞k→∞k→∞k→∞eyα+β{xn+yn}Âñf4¥ö(^y{) b{xn+yn}Âñf{xn+yn}§¦limxn+yn=γ>α+βk0k0k0→∞k0k0 Klimyn=limxn+yn−limxn=γ−α>βk0→∞k0k0→∞k0k0k0→∞k0ùβ{yn}¤kÂñf4¥gñ.u´α+βÒ´{xn+yn}¤kÂñf4.Óny§α>0§α+β{xn+yn}Âñf4¥.llim(xn+yn)=α+β=limxn+limynn→∞n→∞n→∞lim(xn·yn)=αβ=limxn·limyn§elimxn>0n→∞n→∞n→∞n→∞4.¦eêþ4e4µ1(1)an=(n=1,2,···)2−n+(−1)n!1n(2)an=(−1)1+(n=1,2,···)nn(−1)(3)an=(n=1,2···)nnπ(4)an=sin(n=1,2,···)5)µ(1)§küä4fêµa2k→1,a2k+1→−1(k→∞)(k=1,2,3···)u´liman=1,liman=−1.n→∞n→∞(2)§küä4fêµa2k→1,a2k+1→−1(k→∞)(k=1,2,3···)u´liman=1,liman=−1.n→∞n→∞课后答案网(3)Ïliman=0§liman=0,liman=0.n→∞n→∞n→∞2nπ2(4)−sinπ6sin6sinπ5552n=10k+2(k=1,2,···)§a10k+2→sinπ(k→∞)5www.hackshp.cn2n=10k−2(k=1,2,···)§a10k−2→−sinπ(k→∞)522u´liman=sinπ,liman=−sinπ.n→∞5n→∞5pp5.elimn|ann|=α§Klim|ak|=α0+nn→∞n→∞d?k0´?¿½ê.y²k0111n(1)Ï|ak|n=|ak|k0+n|ak|k0+n0+n0+n0+npk0k011nk0+n|ak0+nk0+nqlimk|=α§α>0§limln|ak|=0§lim|ak|=10+n0+n0+nn→∞pn→∞nn→∞d12K(1)§limn|ak|6α0+nn→∞p1nn(2)Ïlim|an|=α§3f{an}§¦lim|an|k=α§kkn→∞k→∞k0111nk−k0α>0§klim|an|nk−k0=lim|an|nk·lim|an|nk=αkkkk→∞k→∞k→∞若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn144pllimn|ak|>α0+nn→∞nÜ(1)(2)§α>0§(ؤá.ppk0+n1nn|ank0+n(3)eα=0§Kw,klimn|=0§llim|ak|=lim|ak|=00+n0+nn→∞n→∞n→∞u´d(Ø(.6.eliman=aN§kanN§kan1(n=1,2,···)n→∞课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn145§2.?êÂñ59ÙÄ 51.?Øe?êñÑ5µ111(1)++···++···1·66·11(5n−4)(5n+1)23n(2)1+++···++···352n−1!!!111111(3)++++···+++···2322322n3n111(4)++···++···1·44·7(3n−2)(3n+1)πππ(5)cos+cos+cos+···345)µ"#!11111111111(1)ÏSn=++···+=1−+−+···+−=1−1·66·11(5n−4)(5n+1)566115n−45n+155n+1!111KlimSn=lim1−=n→∞n→∞55n+15111u´â½Â§?ê++···++···Âñ.1·66·11(5n−4)(5n+1)n1(2)Ïlim=6=0§?êuÑ.n→∞2n−12X∞1X∞1(3)duþÂñAÛ?ê§2n3nn=1n=1!X∞11X∞1X∞113dê?ê52§+=+=1+=2n3n2n2n22n=1n=1n=1"#!11111111111(4)ÏSn=++···+=1−+−+···+−=1−1·44·7(3n−2)(3n+1)34473n−23n+133n+1!111KlimSn=lim1−课后答案网=n→∞n→∞33n+13111u´â½Â§?ê++···++···Âñ.1·44·7(3n−2)(3n+1)π(5)Ïlimcos=16=0§?êuÑ.n→∞n+22.|^ÜÂñnOe?ê´Âñ´uÑ.www.hackshp.cn2n(1)a0+a1q+a2q+···+anq+···,|q|<1,|an|6A,(n=0,1,2,···)11111(2)1+−++−+···23456y²µ(1)Ïé?Ûg,êp§|Sn+p−Sn|=an+an+1qn+1+···+an+p−1qn+p−16a|qn|+aqn+1+···+aqn+p−16nqnn+1n+p−1p1−|q|nA|q|1−|q|n|q|pq|q|<1§K0<1−|q|<1§u´|Sn+p−Sn|0§N=ln/ln|q|§n>N§é?Ûp=1,2,3,···§Ao¤á|Sn+p−Sn|<ε2nUÂñn§?êa0+a1q+a2q+···+anq+···Âñ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn146!X∞111(2)d?ê+−3n+13n+23n+3n=010<ε0<§Ønõ§e-p=n§Kk611111111|Sn+p−Sn|=|S2n−Sn|=+−+···++−>+−3n+13n+23n+36n−26n−16n3n+33n+3!1111111111111+···++−=++···+>++···+=>ε03n+36n6n6n3n+1n+22n32n2n2n6|{z}!nX∞111Ïd?ê+−uÑ.3n+13n+23n+3n=0X∞X∞3.k?êan(=zan>0)§Áy²eéÙ)Ò ¤|¤?êÂñ§Kan½Âñ.n=1n=1X∞X∞y²µanÜ©Úê{Sn}§)Ò ¤|¤?êAnn=1n=1X∞Ù¥An=ai+ai+···+ai§w,AnE?ê.n−1+1n−1+2nn=100ÙÜ©Úê{Sn}§Ù¥Sn=(a1+a2+···+ai)+(ai+···+ai)+···+(ai+···+ai)11+12n−1+1n0w,Sn>SnX∞000qAnÂñ§dÄ ½n§{Sn}kþ.§=3M>0§¦Sn6M§lSn6Sn6M§`n=1²{Sn}kþ.X∞KdÄ ½n§anÂñ.n=14.(½¦e?êÂñx.X∞1(1)(1+x)nn=0X∞n(2)(lnx)n=1)µ课后答案网11(1)d?ê´ú" "?ê§<1?êÂñ1+x1+xlÂñx<−2½x>0.(2)d?ê´ú"lnx "?ê§|lnx|<1?êÂñ1lÂñ1)1+ann=1X∞1(6)√n·nnn=1!nX∞1(7)2n+1n=1X∞1(8)[ln(n+1)]nn=1X∞2+(−1)n(9)2nn=1X∞πn(10)2sin3nn=1X∞nn(11)n!n=1X∞xn(12),(x>0)(1+x)(1+x2)···(1+xn)n=1!n课后答案网X∞b(13)§Ù¥an→a,an,b,aê§a6=0ann=1)µ1√X∞n2+nn1(1)Ïlim=lim√www.hackshp.cn=1§?ê´uÑn→∞1n→∞n2+nnn=1nX∞1Kd"O{§?ê√½uÑ.n2+nn=11un+1(2n+1)22n+12n−11(2)Ïlim=lim=lim=<1n→∞unn→∞1n→∞4(2n+1)4(2n−1)22n−1X∞1KdKO{§?êÂñ.(2n−1)·22n−1n=1√∞√n−n1Xn−n(3)Ïlim=90§?êuÑ.n→∞2n−122n−1n=1ππX∞πX∞π(4)Ïsin6§Âñ§?êsinÂñ.2n2n2n2nn=1n=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn148!n!n11X∞1X∞1(5)Ï6§Âñ§?êÂñ.1+anaa1+ann=1n=1!1nxlimxlnx11(6)Ïlimxx=limelnx=limexlnx=ex→+0=1§lim√=lim=1x→+0x→+0x→+0n→∞nnn→∞n1√∞n·nn1X1qlim=lim√=1§?ê´uÑn→∞1n→∞nnnn=1nX∞1Kd"O{§?ê√uÑ.n·nnn=1vu!n!nu11X∞1(7)Ïlimtn=lim=0<1§?êÂñ.n→∞2n+1n→∞2n+12n+1n=1s11X∞1n(8)Ïlim=lim=0<1§?êÂñ.n→∞[ln(n+1)]nn→∞ln(n+1)[ln(n+1)]nn=12+(−1)n3X∞3(9)Ï6?êÂñ2n2n2nn=1X∞2+(−1)nKâ"O{§?êÂñ.2nn=1!n!nπ2X∞2n(10)Ï0<2sin6π?êÂñ3n33n=1X∞πnKâ"O{§?ê2sinÂñ.3nn=1n+1(n+1)!nX∞nun+1(n+1)!1n(11)Ïlim=lim=lim1+=e>1§?êuÑ.nn→∞unn→∞nn→∞nn!n=1n!n+12nn+10<1,x>1½x=0un+1x/[(1+x)(1+x)···(1+x)(1+x)]x1(12)Ïlim=lim=lim=<1,x=1n→∞unn→∞课后答案网xn/[(1+x)(1+x2)···(1+xn)]n→∞1+xn+12x<1,01=b>a§?êuѶaann=1!n!nbX∞1X∞1X∞1=1=b=a§I?Úä"~Xµ?ê√=uѶ?ê√=annnnn2n=1n=1n=1X∞1Âñ.n2n=1X∞X∞22.e?êunÂñ§y²unÂñ§Ù_XÛºn=1n=1X∞y²µÏunÂñ§Klimun=0n→∞n=12ε0=1§K3êN§n>N§k|un|<ε0=1=06un<1§u´06unN)§若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn149X∞2ld"O{§unÂñ.n=1X∞1X∞1X∞1X∞1Ù_Øý.~µÂñ§uѶÂñ§Âñ.n2nn4n2n=1n=1n=1n=1X∞X∞unX∞X∞X∞X∞3.unÚvnü?ê§lim=0§y²µvnÂñ§unÂñ.qevnuѧunXn→∞vnn=1n=1n=1n=1n=1n=1unX∞X∞Ûºelim=∞§@ounÚvnñÑ5mko"Xºn→∞vnn=1n=1y²µunX∞X∞(1)Ïlim=0§unÚvnü?ên→∞vnn=1n=1ununε0=1§K3êN§n>N§k<ε0=1=06<1§u´unN)vnvnX∞X∞qvnÂñ§Kd"O{§unÂñn=1n=1X∞X∞evnuѧKunUÂñ§UuÑn=1n=11X∞1X∞1n2~µuѧlim=0§Âñ¶nn→∞1n2n=1n=1n1X∞1X∞1√uѧlimn=0§uÑ.nn→∞1nn=1√n=1nunX∞X∞(2)Ïlim=∞§unÚvnü?ên→∞vnn=1n=1unG0=1§K3êN§n>N§k>G0=1§u´un>vn(n>N)vnX∞X∞X∞X∞X∞X∞eunÂñ§Kd"O{§vnÂñ¶evnuѧKunuѶeunuѧKvnñn=1课后答案网n=1n=1n=1n=1n=1Ñ5ؽ.X∞X∞X∞X∞4.eü?êunÚvnuѧmax(un,vn)§min(un,vn)ü?êXÛºn=1n=1n=1n=1X∞X∞)µÏü?êunÚvnuѧun6max(un,vn)n=1n=1X∞www.hackshp.cnKd"O{§max(un,vn)uÑ.n=1X∞éumin(un,vn)ñÑ5ؽ.n=1!X∞1X∞1X∞11X∞1~µ§Ñuѧmin,=uѶn2nn2n2nn=1n=1n=1n=1!X∞1+(−1)nX∞1−(−1)nX∞1+(−1)n1−(−1)n§Ñuѧmin,=0+0+···+0+···%Âñ.2222n=1n=1n=15.|^?êÂñ7^y²µnn(1)lim=0n→∞(n!)2(2n)!(2)lim=0(a>1)n→∞an!y²µ若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn150X∞X∞nn(1)un=(n!)2n=1n=1n+1(n+1)!nun+1[(n+1)!]211Ïlim=lim=lim1+=0<1nn→∞unn→∞nn→∞n+1n(n!)2X∞nnnnKâKO{4/ª§Âñ§ld?êÂñ7^§lim=0(n!)2n→∞(n!)2n=1X∞X∞(2n)!(2)un=an!n=1n=1[2(n+1)]!2un+1a(n+1)!(2n+2)(2n+1)4(n+1)Ï0<==<(a>1)un2(n−1)!n+1n(2n)!aaan!k2n4(n+1)un+1lim=0(a>1,k∈N)§u´lim=0§llim=0n→∞ann→∞an+1n→∞unX∞(2n)!(2n)!KâKO{4/ª§Âñ§ld?êÂñ7^§lim=0an!n→∞an!n=16.?Øe?êÂñ5µX∞1(1)n·(lnn)pn=1X∞1(2)n·lnn·lnlnnn=1X∞1(3)n·(lnn)1+σlnlnnn=1X∞1(4)n·(lnn)p(lnlnn)qn=1)µ1(1)duØpÛê§x¿©§¼êf(x)=Ñ´K4~§课后答案网x(lnx)pZndx11−p(ln2),p>1limp=p−1n→∞2x(lnx)∞,p61p>1§?êÂñ¶p61§?êuÑ.1(2)f(x)=§f(x)x>3´4~¼ê.Zxlnxlnlnxwww.hackshp.cn∞ndxX1lim=lim(lnlnlnn−lnlnln2)=∞§K?êuÑ.n→∞3xlnxlnlnxn→∞n·lnn·lnlnnn=1Z!ndx1111(3)Ïlim=lim−=(σ>0)n→∞2x(lnx)1+σn→∞σ(ln2)σ(lnn)σσ(ln2)σX∞1?êÂñ.n(lnn)1+σn=211X∞1q6§Kd"O{§?êÂñ.n·(lnn)1+σlnlnnn(lnn)1+σn·(lnn)1+σlnlnnn=11(4)-f(x)=§n63´4~¼ê.x(lnx)p(lnlnx)qZZ+∞dx+∞dtqÏ=3x(lnx)p(lnlnx)qlnln3e(p−1)ttqé?Ûq§p−1>0§È©Âñ§p−1<0§È©uѶp=1§eq>1§È©Âñ§eq61§È©uÑ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn151dÜÈ©O{§?êñÑ5È©ñÑ5^K?êp>1Âñ¶p<1uѶp=1§q>1?êÂñ¶q61?êuÑ.X∞7.eun´Âñ?꧿ê{un}üNeü§y²limnun=0.n→∞n=1X∞X∞Xny²µÏunÂñ§S=un§Sn=ukn=1n=1k=1KlimSn=S=limS2n§u´lim(S2n−Sn)=0n→∞n→∞n→∞q{un}üNeü§KS2n−Sn=un+1+un+2+···+u2n>u2n+u2n+···+u2n=nu2nqun>0§K06nu2n6S2n−Sn§u´limnu2n=0§llim(2n)u2n=0n→∞n→∞2n+1qÏu2n+16u2n,un>0§K06(2n+1)u2n+16(2n+1)u2n=(2nu2n)→0(n→∞)2nu´lim(2n+1)u2n+1=0§llimnun=0n→∞n→∞8.y²KO{9Ù4/ª.y²µ(1)KO{µun+1uN+2uN+3uN+k+1Ïn>N§k6q<1§K6q,uN+26quN+1;6q,uN+36quN+2;···;6unuN+1uN+2uN+kKq,uN+k+16quN+k6···6quN+1X∞X∞kkÏq<1§KqÂñ§u´dÂñ?ê51§quN+1Âñ§ld"O{§k=1k=1X∞uN+kÂñk=2X∞2dÂñ?ê55§Vku1,u2,···,uN+1 #?êunÂñ.n=1un+1uN+1en>N§>1§K>1,uN+1>uN§ù`²{uN}´üNOunuNX∞qun>0§Kun90(n→∞)§u´unuÑ.n=1(2)KO{4/ªµun(i)elim=¯r<1n→∞un−1课后答案网d¢êÈ573ε0>0(§¦r<)r¯+ε0<1un+1dþ4½n1y²¥§kkur¯+ε0§u´½3êN(unun+1k¥eIN=)§¦n>N§k1n→∞un−1d¢êÈ573ε0>0(§¦r>r¯)−ε0>1un+1dþ4½n2y²¥§kkur+ε0§u´½3êN(unun+1k¥eIN=)§¦n>N§k>r+ε0>1§dKO{un?êuÑ.X∞X∞1un+1un+1X∞X∞1(iii)Þ~`²µun=§lim=lim=1§un=uѶnn→∞unn→∞unnn=1n=1n=1n=1X∞X∞1un+1un+1X∞X∞1un=§lim=lim=1§un=Âñ.n2n→∞unn→∞unn2n=1n=1n=1n=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn152§4.?¿?ê1.?Øe?êÂñ5£)^Âñ½ýéÂñ¤µ131313(1)−+−+−+···21022103231051111(2)1−+−+−···23!45!X∞lnnn−1(3)(−1)nn=2X∞n3n−1(4)(−1)2nn=1X∞nn+1(5)(−1)(n+1)2n=1X∞xn(6)(−1)sin(x6=0)nn=1111111(7)√−√+√−√+···+√−√+···2−12+13−13+1n−1n+1)µX∞1X∞1X∞3(1)ÏÂñ§Âñ§KÂñ2n102n−1102n−1n=1n=1n=1X∞1X∞1131313u´+Âñ§=++++++···Âñ§l?êýéÂñ.2n102n−12102210323105n=1n=1!X∞1X∞1(2)ÏuѧK−uÑn2nn=1n=1X∞1qé?ê(2n−1)!n=11X∞(2n+1)!11Ïlim=lim=0<1§KdKO{4/ª§?êÂñ1n→∞n→∞2n(2n+1)(2n1)!(2n−1)!课后答案网n=1u´?êuÑ.X∞lnnX∞lnnn−1(3)Ï(−1)=nnn=2n=2lnnX∞1X∞lnnnqlim=limlnn=+∞uѧKd"O{§uÑ1n→∞n→∞nnnn=1!n=2www.hackshp.cn0lnxlnx1−lnxlnx0qf(x)=(x>3)§Kf(x)==<0(x>3)§u´f(x)=üNeü§lxxx2x()lnn3n>3üNeünlnx1lnnqlim=lim=0§Klim=0x→+∞xx→+∞xn→∞nX∞lnnn−1u´â4ÙZ[½n§(−1)^Âñ.nn=2X∞n3X∞n3n−1(4)(−1)=2n2nn=1n=13!3(n+1)1n+11X∞n32n+1Ïlim=lim=<1§KâKO{§Âñn→∞n3n→∞2n22n2nn=1X∞n3n−1l(−1)ýéÂñ.2nn=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn153X∞nX∞nn+1(5)(−1)=(n+1)2(n+1)2n=1n=1n2X∞X∞(n+1)2n1nÏlim=lim=1uѧKuÑn→∞1n→∞(n+1)2n(n+1)2nn=1n=1x1−x0f(x)=(x>2)§Kf(x)=<0(x>2)§u´x>2§f(x)üNeü§l((x)+1)2(x+1)3nn>2üNeü(n+1)2nX∞nn+1qlim=0§Kâ4ÙZ[½n§(−1)Âñn→∞(n+1)2(n+1)2n=1X∞nn+1l(−1)^Âñ.(n+1)2n=1X∞xX∞xn(6)(−1)sin=sinnnn=1n=1sinxX∞1X∞xnÏ→|x|6=0(n→∞)uѧKsinuÑ1nnnn=1n=1xxπ+qé∀x∈R,x6=0§Ï→0(n→∞)§K3N∈Z§n>N§k0<<§u´n>nn2xxX∞xnN§sinxkÓÎÒsinnO~0§Kd4ÙZ[O{§(−1)sinÂnnnn=1ñX∞xnl(−1)sin^Âñ.nn=1!nX+111nX+12Xn1(7)Ü©Úê{Sn}§KS2n=√−√==2k−1k+1k−1kk=2k=2k=1u´limS2n=+∞§Kd?ê)Ò uѧl?êuÑ.n→∞2.y²µe?ê)Ò ¤¤?êÂñ§¿3Ó)ÒSÎÒÓ§@"K)Ò §d?√X∞(−1)[n]ê½Âñ¶¿d?êÂñ5.nn=1课后答案网y²:X∞0(1)®#?êun=(u1+···+un1)+(un1+1+···+un2)+···+(unk−1+1+···+unk)+···Âñn=13Ó)ÒSÎÒÓXnXn00000uk=Sn,uk=Sn§KSwww.hackshp.cn1=Sn1,S2=Sn2,···,Sk=Snk,···k=1k=1X∞?êeInlnk−1nk§unÜ©ÚüNCz§=n=100un,···,unþ§kSk−1=SnSn>Sn=Skk−1+1kk−1kX∞X∞0Âñ§=limS0000®unk=limSk−1=S§KlimSn=S§u´unÂñ.k→∞k→∞n→∞n=1n=1√X∞(−1)[n](2)Änn=1111X∞222kn=k,k+1,···,k+2k(k=1,2,···)§ÃanÓÒ.PAk=++···+§(−1)Ak´k2k2+1k2+2kk=1?êZk2+2kZ(k+1)222dx111dxk+2k(k+1)Ï6++···+6=ln6Ak6ln§lk2−1xk2k2+1k2+2kk2xk2−1k2k→∞§Ak→0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn154k2+2k(k+2)2k2+kX∞kqAk−Ak+1>ln−ln=ln>0§Kd4ÙZ[O{(−1)AkÂk2−1(k+1)2k2+k−2k=1ñ§l?êÂñ.3.?Øe?ê´ÄýéÂñ½^ÂñµX∞(−1)n(1)n+xn=1X∞sin(2nx)(2)n!n=1X∞sinnx(3),(00§üN~§lim=0§Kd4ÙZ[½n§Âñn+xn→∞n+xn+xn=1+x<0ØKê§Ïx½ê§Kn¿©§=3N∈Z§n>N§kn+x>X∞(−1)n11X∞(−1)n0§u´´?ê§düN~9lim=0§KÂñ§ln+xn+xn→∞n+xn+xn=N+1n=N+1X∞(−1)nÂñn+xn=1KxØKê§d?ê^Âñ.sin(2nx)111X∞1(n+1)!(2)Ï6§lim=lim=0<1§KdKO{§Âñ1n!n!n→∞n→∞n+1n!n!n=1X∞sin(2nx)X∞sin(2nx)2â"O{§Âñ§lýéÂñ.n!n!n=1n=1课后答案网()Xncosx−cos2n+1x11(3)Ïsinkx=226êüNªu02sinxsinxnk=122X∞sinnxKd)á4O{§Âñ.nn=1()sinnxsin2nx1cos2nxXnsinx−sin(2n+1)x11q>=−www.hackshp.cncos2kx=69êünn2n2n2sinx|sinx|2nk=1Nªu0X∞cos2nxKd)á4O{§Âñ.nn=1!X∞1X∞1X∞1cos2nxX∞sinnxquѧKuѧu´−uѧluÑn2n2n2nnn=1n=1n=1n=1X∞sinnxK?ê(01§Ï6p>1Âñ§K?ê(0=+dfây²Âñ.npnp2np2np(2n)pn=1X∞cos2nxX∞cos2nxp−1K·2Âñ§=Âñ(2n)p2npn=1n=1!X∞1X∞1X∞1cos2nxq0x0Âñ.nx0课后答案网nxn=1n=11!x−x0!x−x0(n+1)x−x0n111y²µÏx>x0§K==1−<1§K<1n+1n+1(n+1)x−x0nx−x0nx−x0161nx−x0()11u´êüNk.§61nx−x0nwww.hackshp.cnx−x0X∞anX∞anq?êÂñ§KdCO{§Âñ.nx0nxn=1n=1X∞X∞6.{nan}Âñ§n(an−an−1)Âñ§KanÂñ.n=1n=1y²µÏ{nan}Âñ§Ù4a()X∞Xnqn(an−an−1)Âñ§KÙÜ©Úêk(ak−ak−1)k4§Ù4Sn=1k=1XnnX−1qk(ak−ak−1)=(a1−a0)+2(a2−a1)+···+n(an−an−1)=nan−akk=1k=0nX−1XnnX−1Xn=ak=nan−k(ak−ak−1)§Klimak=limnan−limk(ak−ak−1)=a−Sn→∞n→∞n→∞k=0k=1k=0k=1X∞X∞X∞u´anÂñ§lan=an−a0Âñ.n=0n=1n=0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn156X∞X∞X∞7.e(av−av−1)ýéÂñ§bvÂñ§@"avbvÂñ.v=1v=1v=1nX+mn+my²µ-Bn=bvv=n+1nX+pXp−1n+pn+idAbelC§avbv=an+pBn+Bn(an+i−an+i+1)v=n+1i=1nX+pXp−1avbv6|an+p|Bn+p+Bn+i|annn+i−an+i+1|v=n+1i=1"#nX+pXp−1-Hp=maxBn+1,Bn+2,···,Bn+p§Kkavbv6Hp|an+p|+|an+i−an+i+1|nnnnnv=n+1i=1X∞X∞X∞Ï|av−av−1|Âñ§(av−av−1)Âñ(av−av−1)=−a0+an§liman3n→∞v=1v=1v=1Ï3M>0§¦én§kXp−1|an+i−an+i+1|+|an+p|0,∃N∈Z§n>N§ép∈Z§kv=1εpHn<(5)MnX+pX∞d(??),(??)§n>N§kavbv<ε§ùL²?êavbvÂñv=n+1v=18.|^ÜÂñny²?ê4ÙZ[½n.y²µé?Ûg,êp§k|Sn+p−Sn|=(−1)n+2un+1+(−1)n+3un+2+···+(−1)n+p+1un+p=(−1)n+2(un+1−un+2+···+(−1)p−1un+p)=up−1un+1−un+2+···+(−1)n+ppóê§(un+1−un+2)+···+(un+p−1−un+p)>0pÛê§(un+1−un+2)+···+(un+p−2−un+p−1)+un+p>0oup−1un+p=up−1un+pn+1−un+2+···+(−1)课后答案网n+1−un+2+···+(−1)qpóê§un+1−(un+2−un+3)−···−(un+p−2−un+p−1)−un+p6un+1pÛê§un+1−(un+2−un+3)−···−(un+p−1−un+p)6un+1p−1oun+1−un+2+···+(−1)un+p6un+1é?¿ε>0§Ïlimun=0§Klimun+1=0n→∞n→∞+u´73N∈Z§n>N§k|un+1−0|<ε§Kun+1<εpdn>N§é?Ûg,êpÑk|Sn+p−Sn|=un+1−un+2+···+(−1)un+p6un+1<εX∞www.hackshp.cnn+1ldÜÂñn§(−1)unÂñ.n=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn157§5.ýéÂñ?êÚ^Âñ?ê5X∞1v−1v−2v−11.|x|<1,|y|<1§y²(x+xy+···+y)=(1−x)(1−y)v=1y²µÏ|x|<1,|y|<1§KX∞1v−12vx=1+x+x+···+x+···=ýéÂñ(6)1−xv=1X∞1v−12vy=1+y+y+···+y+···=ýéÂñ(7)1−yv=1X∞X∞1v−1v−1(??)·(??)§xy=(1−x)(1−y)v=1v=1X∞X∞X∞v−1v−12v2vv−1v−2v−1qxy=(1+x+x+···+x+···)(1+y+y+···+y+···)=(x+xy+···+y)§v=1v=1v=1X∞1v−1v−2v−1K(x+xy+···+y)=.(1−x)(1−y)v=1X∞xnX∞ynX∞(x+y)n2.y²µ=n!n!n!n=0n=0n=0|x|n+1|x|X∞|x|n(n+1)!y²µÏlimn=lim=0<1§KâKO{4/ª§?êÂñn→∞|x|n→∞n+1n!n!n=0X∞|x|nu´?êýéÂñn!n=0X∞|y|nÓn§?êýéÂñn!n=0X∞xnX∞ynX∞¤=Cnn!n!n=0n=0n=0Xnxiyn−iynxyn−1xn1(x+y)n0n1n−1nnÙ¥Cn=·=+·+···+=(Cny+Cnxy+···+Cnx)=i!(n−i)!n!1!(n−1)!n!n!n!i=0X∞xnX∞ynX∞(x+y)n课后答案网K=n!n!n!n=0n=0n=03.y²µ±Ñ^Âñ?êS?ꧦÙuÑ+∞.X∞y²µun^Âñn=1X∞X∞X∞X∞d½n1§vnÚwnÑuѧwww.hackshp.cnvnuÑ+∞§(−wn)uÑ−∞n=1n=1n=1n=1ÀuÑ+∞ê{βn}§=limβn=+∞n→∞X∞rvnU^Så5n=1m1§¦v1+v2+···+vm>β1+w11, m2§¦v1+v2+···+vm+vm+···+vm>β2+w1+w211+12/§¿©mk>mk−1§¦v1+v2+···+vm+···+vm+···+vm>βk+w1+w2+···+wk(k=12k3,4,···)ù/|ÚKµ(v1+···+vm−w1)+(vm+···+vm−w2)+···+(vm+···+vm−wk)+···(∗)11+12k−1+1kd?êw,?êS?êX∞Ï(∗))Ò ?ê(vm+···+vm−wk)kgÜ©Úk−1+1kk=1(v1+···+vm−w1)+(vm+···+vm−w2)+···+(vm+···+vm−wk)>βk11+12k−1+1k若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn158limβk=+∞k→∞X∞K(vm+···+vm−wk)uÑ+∞k−1+1kk=1duÑ?ê?¿)Ò§K±Ñ^Âñ?êS?ꧦÙuÑ+∞.课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn159§6.á¦È1.?Øá¦ÈÂñ5µY∞n2−4(1)n2−1n=3Y∞(−1)n(2)an(a>0)n=1sY∞n+1(3)n+2n=0)µ2n−433(1)Ï=1−,n>3§−<0n2−1n2−1n2−1!3X∞1X∞3X∞3n2−1qlim=3Âñ§KÂñ§u´−Âñn→∞1n2n2−1n2−1n2n=3n=3n=3Y∞n2−4lâ½n2§Âñ.n2−1n=3X∞(−1)nX∞(−1)nX∞(−1)n(2)lnan=lna=lnannn=1n=1n=1X∞(−1)nX∞(−1)nY∞(−1)nÏ?ê4ÙZ[.?ê§KÙÂñ§u´?êlnanÂñ§lá¦ÈanÂnn=1n=1n=1ñ.ss12nn+11(3)duÜ©¦ÈPn=·····=→0(n→∞)23n+1n+2n+2sY∞n+1á¦ÈuÑu0.n+2n=0X∞Y∞22.y²µexnÂñ§KcosxnÂñ.n=1n=1!!2Y∞Y∞xnxnsinxnx222ny²µÏcosxn=1−2sin课后答案网062sin62·=2222n=1n=1X∞X∞xn22qxnÂñ§K2sinÂñ2n=1n=1Y∞u´â½n2§cosxnÂñ.n=1!!X∞Y∞www.hackshp.cnππ3.y²µeαnýéÂñ§Ktan+αnÂñÙ¥|αn|<.44n=1!n=1!Y∞πY∞1+tanαnY∞2tanαny²µtan+αn==1+41−tanαn1−tanαnn=1n=1n=1∞2tanαnX1−tanαn2tanαnÏαnýéÂñ§Klimαn=0§u´lim=lim=2n→∞n→∞|αn|n→∞1−tanαnαnn=1X∞X∞X∞2tanαn2tanαndαnýéÂñ§Âñ§u´ýéÂñ1−tanαn1−tanαnn=1!n=1n=1Y∞πltan+αnýéÂñ.4n=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn1601Ù2ÂÈ©§1.á2ÂÈ©1.¦e2ÂÈ©µZ+∞1(1)dxx2−12Z+∞1(2)dx,(p,q>0)0(x2+p)(x2+q)Z+∞−ax2(3)exdx(a>0)0Z+∞−ax(4)esinbxdx,(a>0)0)µZ+∞11x−1+∞1√(1)dx=ln=ln3=ln32x2−12x+122Z!+∞+∞111x1xππ2(2)dx=√arctan√−√arctan√=√√√=√√0(x2+p)(x2+q)q−pppqqpq(p+q)2(qp+pq)0Z+∞−ax2+∞−ax2e1(3)exdx=−=02a2a0Z+∞+∞−ax−asinbx−bcosbx−axb(4)esinbxdx=e=0a2+b2a2+b202.?ØeÈ©Âñ5µZ+∞dx(1)√3x4+10Z+∞xarctanx(2)dx1+x2课后答案网1Z+∞1(3)sindxx21Z+∞dx(4)01+x|sinx|Z+∞xwww.hackshp.cn(5)dx1+x2sin2x0Z+∞xm(6)dx,(n>0,m>0)1+xn0)µZZZ+∞dx1dx+∞dx(1)√=√+√3x4+13x4+13x4+10Z011dxÏ√~È©§KÙ7Âñ3x4+1Z0ZZZ+∞dx+∞dx+∞dx+∞dxé√§Ï√Âñ§K√Âñ§l√Âñ.3x4+13x43x4+13x4+11110xarctanxx3πZ+∞11+x3(2)Ïlim=limarctanx=§Âñx→+∞1x→+∞1+x32x2x21Z+∞xarctanxKd"O{4/ª§dxÂñ.1+x21若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn1690ε→0)ZZbCbqp=1§dxuѧl|f(x)|dxuÑ.ax−aa(2)ÜO{4/ªµp(i)lim(x−a)|f(x)|=k(0ε>0§3δ>0§¦a1§f(x)k½Ò§Kf(x)dxuÑaa(ii)k=0§ε0=1§K∃δ>0§¦a0§¦a11=|f(x)|>(x−a)pZZbbKdÜO{§p>1§|f(x)|dxuѶqf(x)k½Ò§lf(x)dxuÑ.aZabnþ§e06k<+∞,p<1§@"f(x)dxýéÂñ¶e01§@Zab"f(x)dxuÑ.a4.?ØeÈ©Âñ5µZ+∞dx(1)p3(x−1)2x(x−2)0Z+∞ln(1+x)(2)dxxα0Z+∞dx(3)课后答案网xp+xq0Z+∞arctanx(4)dxxα0Z+∞dx(5)pq1xlnxZ+∞dxwww.hackshp.cn(6)−∞|x−a1|p1|x−a2|p2···|x−an|pn)µ(1)x=0,1,2þȼêÛ:ZZ1ZZ3ZZZ!+∞123+∞dx22dxp=+++++p3(x−1)2x(x−2)133(x−1)2x(x−2)0012322Z12dxéÈ©p3(x−1)2x(x−2)011x32Ïlimx3p=lim=0x→+03(x−1)2x(x−2)x→+0(x−1)2(x−2)Z12dxKdÜO{4/ª§È©pýéÂñ3(x−1)2x(x−2)0Z1dxéÈ©p13(x−1)2x(x−2)2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn170151(1−x)6Ïlim(1−x)6p=lim1=0x→1−03(x−1)2x(x−2)x→1−0[x(x−2)]3Z1dxKdÜO{4/ª§È©pýéÂñ13(x−1)2x(x−2)2Z32dxÓdy{§È©pýéÂñ3(x−1)2x(x−2)1Z2dxéÈ©p33(x−1)2x(x−2)2112−x32Ïlim(2−x)3p=lim=0x→2−03(x−1)2x(x−2)x→2−0x(x−1)2Z2dxKdÜO{4/ª§È©pýéÂñ33(x−1)2x(x−2)Z23dxÓdy{§pýéÂñ3(x−1)2x(x−2)2Z11+∞dxÏx>3§p6ýéÂñ3244(x−1)x(x−2)(x−2)33(x−2)3Z+∞dxKd"O{§pýéÂñ3(x−1)2x(x−2)3Z+∞dxnþ§pýéÂñ3(x−1)2x(x−2)0ZZZ+∞ln(1+x)1ln(1+x)+∞ln(1+x)(2)dx=dx+dxxαxαxα0Z011ln(1+x)édxxα00,α60ln(1+x)10,0<α<1Ïlim=1+x§Kα>1§0Û:x→+0xαlim=1,α=1x→+0αxα−1∞,α>1ln(1+x)ln(1+x)α−1qlimx=lim=1x→+0xαx→+0xZZ课后答案网1ln(1+x)1ln(1+x)Kα−1<1=1<α<2§dxýéÂñ¶α>2§dxuѶxαxαZ001ln(1+x)α61§dx~È©§7Âñxα0ZZ1ln(1+x)1ln(1+x)lα<2§dxýéÂñ¶α>2§−dxuÑxαxαZ00+∞ln(1+x)www.hackshp.cnédxxα11ln(1+x)λ1+xλ>1§α−λ>0§Ïlimx=lim=0x→+∞xαx→+∞(α−λ)xα−λ−1Z+∞ln(1+x)Kα−λ>0=α>λ>1§È©dxýéÂñ¶xα1Zln(1+x)+∞ln(1+x)αqlimx=+∞§Kα61§È©dxuÑx→+∞xα1xαZZ+∞ln(1+x)+∞ln(1+x)l1<α<2§dxýéÂñ¶Ù§/§dxÑuÑ.xαxα00ZZZ+∞dx1dx+∞dx(3)=+0xp+xq0xp+xq1xp+xqZ1dxéÈ©§min(p,q)=pxp+xq0ZZ1dx1dxep60§K~È©§Âñ0xp+xq0xp+xq若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn171111ep>0§dulimxp=lim=,p=qx→+0xp+xqx→+01+xq−p21,p6=qZ1dxÈ©=p<1=min(p,q)<1Âñxp+xqZ0+∞dxéÈ©§max(p,q)=qxp+xq1111dulimxq=lim=,p=qx→+∞xp+xqx→+∞x−(q−p)+121,p6=qZ+∞dxÈ©=q>1=max(p,q)>1Âñxp+xqZ1+∞dxKÈ©min(p,q)<1max(p,q)>1Âñ.xp+xq0arctanx1x1−α1+x2(4)0<α61§lim=lim=lim=0x→+0xαx→+0αxα−1x→+0α(1+x2)arctanx−αα60§lim=limxarctanx=0x→+0xαx→+0Kα61§0ØÛ:arctanxπαqlimx=limarctanx=x→+∞xαx→+∞2Z+∞arctanxKα61§dÜO{4/ª§dxuÑxα01arctanx1+x21α>1§Ïlim=lim=lim=+∞§K0Û:x→+0xαx→+0αxα−1x→+0αxα−1(1+x2)ZZZ+∞arctanx1arctanx+∞arctanxKdx=dx+dxxαxαxαZ0011arctanxédx§α>1§0Û:xα0arctanxarctanxα−1Ïlimx=lim=1x→+0xαx→+0xKdÜO{4/ª§α−1<1=α<2È©Âñ¶α>2§È©uÑZ+∞arctanxédxxα1ZZπarctanxπ+∞π+∞π42课后答案网24Ï66dxα>1È©Âñ¶dxα61È©uÑxαxαxα1xα1xαZZ+∞arctanx+∞arctanxKd"O{§α>1§dxÂñ¶α61§dxuÑxαxαZ11+∞arctanxo§1<α<2§dxÂñ¶Ù{/dÈ©þuÑ.xα0ZZZ+∞dx2dx+∞dx(5)xplnqx=xplnqxwww.hackshp.cn+xplnqx1Z122dxÄpq§é?¿p1x"lnx#!qqq11(x−1)(x−1)x−1qdux→lim1+0(x−1)xplnqx=limx→1+0xp·lnqx=limx→1+0lnqx=x→lim1+0lnx!q1=lim=11x→1+0xZ2dxKÈ©pq=q<1p?¿Âñ1xlnxZ+∞dxÄpq§ep>1§α>0¿©§¦p−α>1§Ké?¿q§2xlnx!11p−αdux→lim+∞xxplnqx=limx→+∞xαlnqx=0Z+∞dxu´È©pqp>1q?¿Âñ¶2xlnx若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn173Z+∞xpsinxo§p>−2,q>p+1§dxýéÂñ1+xqZ0+∞xpsinxÄdx1+xq1ZAxpq>p§sinxdx62§üN~ªu0(x→+∞)1+xq1Z+∞xpsinxKd)á4O{§dxÂñ1+xq1ppxxq6p§q=p§→1(x→+∞)¶q1+xq3pππx1+u´é∀A>1§7∃N∈Z§¦2Nπ+>Ax>2Nπ+§ðk>441+xq3√ππZA00xp1ZA002000léA=2Nπ+,A=2Nπ+§ksinxdx>sinxdx=42A01+xq3A06Z+∞xpsinxKdÜn§dxuÑ1+xq1ZZ+∞xpsinx+∞xpsinxnþ¤ã§q>p+1>−1§dxýéÂñ¶p+1>q>p>−2§dx^01+xq01+xqÂñ.0,λ60esinxsin2xsin2x0,0<λ<10,λ<1(2)Ïlim=lim=2cos2x=2,λ=1x→+0xλx→+0xλlim=2,λ=1x→+0λxλ−1∞,λ>1∞,λ>1Kλ>1§0Û:Z+∞esinxsin2xZ1esinxsin2xZ+∞esinxsin2xdx=dx+dxxλxλxλ0Z011esinxsin2xédx§È¼ê§0Û:xλ0sinxsinxesin2xesin2xλ−1Ïlimx=lim=2x→+0xλx→+0xZ1esinxsin2xKdÜO{4/ª§λ课后答案网−1<1=λ<2§dxýéÂñ¶xλZ01esinxsin2xλ>2§dxuÑxλZ0+∞esinxsin2xédxxλ1Zesinxsin2xe+∞esinxsin2xwww.hackshp.cnλ>1§6§KdxýéÂñxλxλ1xλZ+∞esinxsin2xu´dx1<λ<2ýéÂñ¶xλ0!!sinx−12e|sin2x|esin2x1−cos4x11cos4x−1λ61§>=e=−xλxλ2xλ2exλxλZZZsinx+∞dx+∞cos4x+∞esin2xÏuÑ,dxÂñ§KdxuÑ1xλ1xλ1xλZ+∞esinxsin2xédxxλZ1ZAAÏesinxsin2xdx=2esinxsinxdsinx64e111Z+∞esinxsin2xqüN~ªu0§Kâ)á4O{§éλ>0kdxÂñxλxλZ1+∞esinxsin2xu´dx0<λ61^Âñxλ0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn174Z+∞esinxsin2xZ+∞esinxsin2xl1<λ<2dxýéÂñ¶0<λ61dx^Âñ.xλxλ00Z+∞sin x+1Z1sin x+1Z+∞sin x+1xxx(3)dx=dx+dx=I1+I2xnxnxn00Z 0Z 1dt1sinx+1+∞sinx+1xxéI1§-x=,dx=−§KI1=dx=dxtt20xn1x2−nïÄI2 sinx+11xÏ6§Kn>1§I2ýéÂñxnxnZ+∞sin x+1Z+∞sin x+1 1−1x xx2dxÏdx=xnxn1−111x2!ZA1Asinx+11−1dx=cosx+621xx2x"!#011nn−1n−3n−32x1−=nx−(n−2)x=x[nx−(n−2)]x2!11n üN~ªu0(x→+∞)Kn∈(0,1]§x1−üNO=x2xn1−1x2u´d)á4O{§0 =−xnxn2xn2xnZ+∞dxZ+∞cos2 x+1x01§7∃k∈Z§¦2kπ+>Ax>2kπ+§!!661π1−nðkxsinx+>sin2kπ+=x课后答案网62 !ππZA00sinx+1ZA00ππ000xu´§éA=2kπ+,A=2kπ+§kdx>sin2kπ+dx=63A0xnA0612KdÜÂñn§n60§I2uÑéI1§dI2(ا2−n>1=n<1ýéÂñ¶1>2−n>0=16n<2^Âñ¶2−nZ60=n6 2uÑ+∞sinx+1o§xdx0www.hackshp.cn0§K|f(x)|62若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn175u´d"O{§|f(x)|È=²È½ýéÈ.Ø,.Z21Z2dx~µd57~1§1dxÂñ=1ýéÂñ1(x−1)21(x−1)2Z2dx1uѧ=3[1,2]þØÈ.1x−1x−19.OeÈ©Ü̵Z3dx(1)01−xZ+∞(2)sinxdx−∞)µZ"ZZ#"1−η3#3dx1−ηdx3dx(1)P.V.=lim+=lim−ln(1−x)−ln(x−1)=−ln201−xη→001−x1+η1−xη→001+ηZZA!+∞A(2)P.V.sinxdx=limsinxdx=lim−cosx=lim(cos(−A)−cosA)=0A→+∞A→+∞A→+∞−∞−A−A10.y²2ÂÈ©9ÜÌm"XµZZ+∞+∞(1)ef(x)dxÂñ§ÙA§KÜÌP.V.f(x)dx3§ uA§Ø,¶−∞−∞ZZ+∞+∞(2)ef(x)>0§P.V.f(x)dx3§ÙA§Kf(x)dxÂñ§ÂñuA.−∞−∞y²µZZZZ+∞+∞0+∞(1)df(x)dxÂñ§f(x)dx=f(x)dx+f(x)dxÂñ§−∞Z−∞Z−∞0Z0AAKklimf(x)dx+limf(x)dx3§AOB=−A§klimf(x)dx3§ B→−∞A→+∞A→+∞B0−AuAZ+∞ùL²P.V.f(x)dx3§ uA−∞Ø,.ZZZZ+∞课后答案网A+∞+∞~XµP.V.sinxdx=limsinxdx=0§f(x)dx=sinxdxØÂñ.A→+∞−∞−A−∞−∞(2)^y{.ZaZ+∞eØ,§Kduf(x)>0§f(x)dxÚf(x)dx¥k+∞ZZ−∞aaAu´f(x)dxÚf(x)dwww.hackshp.cnx¥A→+∞kªu+∞§,u u0§l§Ú−AaZZ+∞+∞ªu+∞§ù®P.V.f(x)dx3gñ§Kf(x)dxÂñ.Z−∞Z−∞+∞+∞qdP.V.f(x)dx=A§Kâ45§f(x)dx=A.−∞−∞若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn1761Ü©¼ê?ê1Ù¼ê?ê!?ê§1.¼ê?êÂñ1.?Øe¼êS3¤«SÂñ5µr1(1)fn(x)=x2+,−∞ε0222Kfn(x)3(−∞,+∞)þØÂñ.(4)06x<1§f(x)=limfn(x)=0¶x=1§fn(1)=0,f(1)=0§Kf(x)=limfn(x)=0n→∞n→∞nnn+1||fn−f||=sup|fn(x)−f(x)|=sup|x(1−x)|=max(x−x)x∈[0,1]x∈[0,1]x∈[0,1]nnn+10n−1-(x−x)=x[n−(n+1)x]=0"Kx=0,x=!!n!n+1nnnqfn(0)=fn(1)=0,fn=1−>0n+1n+1n+1!n!nnnn+1Kmax(x−x)=1−→0(n→∞)=||fn−f||→0(n→∞)x∈[0,1]n+1n+1u´d½Â2§d¼êS3¤«SÂñu0.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn1771,00§Ïlimtlnt=0§K3δ(ε)>0§0N§<δδnxxxlé00,∃N=§n>N§k|rn(x)|<ε§Kd?ê3(−∞,+∞)þÂñ.εsinnx1X∞1(3)−∞0§ε=1§73N=N(ε)∈Z(§xÃ")§¦n>N§éu(0,+∞)Sx§þk|un+1(x)+un+2(x)+···+un+p(x)|<ε§Ù¥p?¿ê8p=1,n=N§Kéx∈(0,+∞)§Ak|uN+1(x)|<ε=12qx0=∈(0,+∞)§Ak|uN+1(x0)|<13N+1π1N+1N+1¯¢þ%kuN+1(x0)=2sin=2>1ù|uN+1(x0)|<1gñ3N+1x0X∞1nKbؤá§=?ê2sin3(0,+∞)þÂñÂñ.课后答案网3nxn=13.y²Âñ½Â1Ú½Â2 d5.y²µ½Â1=⇒½Â2ε®é?ε>0§36uεêN(ε)§¦n>N(ε)§k|Sn(x)−S(x)|<éx∈XѤ2áεu´||Sn−S||=sup|Sn(x)−S(xwww.hackshp.cn)|6<ε§llim||Sn−S||=0.x∈X2n→∞½Â2=⇒½Â1+®lim||Sn−S||=0§=é∀ε>0,∃N(ε)∈Z§¦n>N§éx∈X§Ñkn→∞||Sn−S||−0=sup|Sn(x)−S(x)|<εx∈X|Sn(x)−S(x)|6||Sn−S||−0<εéx∈XѤá.X∞( aq/y²¼ê?êun½Â1⇐⇒½Â2).n=1X∞ln(1+nx)4.Áy?ê3?Û«m[1+α,∞),α>0Âñ.nxnn=1ln(1+nx)ln(1+nx)nx11y²µÏh>0§ln(1+h)0)þÂñ.(1+α)n−1n=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn179X∞X∞X∞5.eun(x)|un(x)|6cn(x)§¿cn(x)3XþÂñ§Kun(x)3Xþ½Âñýén=1n=1n=1Âñ.X∞y²µÏcn(x)3XþÂñn=1+KdÂñÜ¿^§é∀ε>0,∃N=N(ε)∈Z§¦n>N§éx∈XÚ?¿êp§k|cn+1(x)+cn+2(x)+···+cn+p(x)|<εX∞qun(x)|un(x)|6cn(x)n=1Kéþãε>0§êN=N(ε)§¦n>N§éx∈XÚþãêp§k|un+1(x)+un+2(x)+···+un+p(x)|6|un+1(x)|+|un+2(x)|+···+|un+p(x)|6|cn+1(x)+cn+2(x)+···+cn+p(x)|<εX∞dÂñÜ¿^§un(x)3XþÂñýéÂñ.n=1Zx6.f0(x)3[0,a]þëY§qfn(x)=fn−1(t)dt§y²{fn(x)}3[0,a]þÂñu".0y²µÏf0Z(x)3[0,a]þëY§KÙk.§=3M>0§k|f0(x)|6Mxqfn(x)=fn−1(t)dt§KZ0ZZxxx|f1(x)=f0(t)dt6|f0(t)|dt6Mdt=Mx6MaZ0Z0Z0xxxMx2Ma2|f2(x)=f1(t)dt6|f1(t)|dt6Mtdt=6.......................................................00022ZxZxZxtn−1xnan|fn(x)=fn−1(t)dt6|fn−1(t)|dt6Mdt=M6M000(n−1)!n!n!ananan+Ïlim=0=é∀ε>0,∃N∈Z§n>N§k−0=<εn→∞n!n!n!u´|fn(x)−0|0§N=§Kn>N§éx∈(−∞,+∞)§Ñk−0=6<εεn+x2n+x2n()www.hackshp.cn1X∞1n−1K"ux∈(−∞,+∞)Âñu0§u´d)á4O{§(−1)3(−∞,+∞)Sn+x2n+x2n=1Âñ.X∞1X∞1n−1(−1)=n+x2n+x2n=1n=11X∞1X∞1n+x2Ïlim=1uѧKd"O{4/ª§uѧu´é?Ûx?êýén→∞1nn+x2nn=1n=1Âñ.X∞x2X∞x2=(1+x2)n(1+x2)nn=1n=1s21xn,x6=0é½x∈(−∞,+∞)§Ïlim2n=1+x2n→∞(1+x)0,x=0X∞x2dÜO{§3(−∞,+∞)Âñ§u´ýéÂñ.(1+x2)nn=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn180Xnx21x6=0§Sn(x)==1−§S(x)=limSn(x)=1(1+x2)k(1+x2)nn→∞k=11,x6=0x=0§Sn(0)=0,S(0)=0§KS(x)=0,x=0X∞x2ÏSn(x)3(−∞,+∞)þëY§S(x)3(−∞,+∞)þØëY§K3(−∞,+∞)SØÂñ.(1+x2)nn=18.y²µXX∞(1)XJ|fn(x)|3[a,b]þÂñ§@"fn(x)3[a,b]þÂñ¶1XXX∞nnn+1(2)XJfn(x)3[a,b]þÂñ§|fn(x)|7Âñ§±(−1)(x−x),06x61~15`².y²µ(1)dÜOK9K§+=é∀ε>0,∃N=N(ε)∈Z§¦n>N§éx∈[a,b]Ú?¿p∈Z§k|fn+1(x)|+|fn+2(x)|+···+|fn+p(x)|<εl|fn+1(x)+fn+2(x)+···+fn+p(x)|6|fn+1(x)|+|fn+2(x)|+···+|fn+p(x)|<εX∞KâÂñÜOK§fn(x)3[a,b]þÂñ.1X∞nnn+1(2)~µ(−1)(x−x)3[0,1]þÂñ1nn+1nn+1nnn+1Ïx−x=0(x=0,1)¶00)(iii)1+ε6x<∞)µπ0,06x6πx6=(1)f(x)=limfn(x)=2n→∞π1,x=2nÏf(x)3[0,π]þØëY§fn(x)3[0,π]þëY§Kfn(x)=(sinx)3[0,π]þØÂñ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn1810,x=0,π(2)(i)f(x)=limfn(x)=n→∞1,00§¦α0§Ke>1§u´lim=<1§KdKO{4/ª§?n→∞ne−nαeαX∞X∞−nα−nxêneÂñ§lâMO{§ne3[α,β]þÂñ.11X∞−nx−nxqne3[α,β]þëY§l课后答案网ne3[α,β]þëY1X∞−nxÏx0∈[α,β]§Kne3x0:ëY1X∞−nxdux0´(0,+∞)?¿:§ne3(0,+∞)SëY.1X∞sinnxwww.hackshp.cn12.y²¼êf(x)=3(−∞,+∞)SëY§¿këY¼ê.n31sinnx1X∞1X∞sinnxy²µÏ6Âñ§KâMO{§f(x)=3(−∞,+∞)Âñn3n3n3n311sinnxX∞sinnxq3(−∞,+∞)SëY§Kf(x)=3(−∞,+∞)SëYn3n3!1dsinnxcosnx=dxn3n2cosnx1X∞1X∞cosnxÏ6Âñ§KâMO{§3(−∞,+∞)Âñn2n2n2n21!1dX∞sinnxX∞cosnx0u´f(x)==dxn3n211cosnxX∞cosnxq3(−∞,+∞)SëY§K3(−∞,+∞)SëYn2n21若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn182X∞cosnx00=f(x)3(−∞,+∞)SëYf(x)=.n21X∞113.y²¼êζ(x)=3(1,+∞)ëY§¿këY ¼ê.nx1X∞lnny²µ¦ê¤?ê−.ey§311?¿5§ζ(x)=−é11)ÑÂñ§ìþã§yµé?Ûnan=1(k)êk§ζ(x)31m§cos(2πy)=1§d?êتu0§Kwww.hackshp.cncos(2πx)=cos(2πy)uѧu´πcos(2πx)u111Ñ−mmq3?Û«mSÑ3x=k+2h(h=0,1,2,···,2−1)ù:§kxêÜ©X∞sin(2nπx)K?ê3?Û«mSØUŦû.2n115.ky1−r2X∞n=1+2rcosnx1−2rcosx+r2n=1|r|<1¤á§ly²µZπ21−rdx=2π(|r|<1)−π1−2rcosx+r2.nny²µ|rcosnx|6|r|é∀x∈(−∞,+∞)ѤáX∞X∞nnÏ|r|<1§K|r|Âñ§u´dMO{§rcosnx3(−∞,+∞)SÂñn=1n=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn183X∞nlf(x)=1+2rcosnxn=122Ï1−2rcosx+r6=0§þªüàÓ¦±1−2rcosx+r§K!X∞22n(1−2rcosx+r)f(x)=(1−2rcosx+r)1+2rcosnx"n=1#X∞X∞X∞2nn+1n+2=1−2rcosx+r+2rcosnx−2r(2cosnxcosx)+2rcosnx"n=1n=1n=1#X∞X∞X∞X∞2nn+1n+1n+2=1−2rcosx+r+2rcosnx−2rcos(n+1)x−2rcos(n−1)x+2rcosnx"n=1n=1!n=1n=1!#X∞X∞X∞X∞2nn+1n+12n+2=1−r+2rcosnx−2rcos(n+1)x+rcosx−2rcos(n−1)x−r+2rcosnxn=1n=1n=1n=12=1−r1−r21−r2X∞nu´f(x)===1+2rcosnx1−2rcosx+r21−2rcosx+r2n=1nduþªmà?ê3[−π,π]þÂñ§rcosnx3[−π,π]þëY§Kþª?ê±ÅÈ©§Zπ2ZπX∞!X∞Zπ1−rnndx=1+2rcosnxdx=2π+2rcosnxdx=2π.−π1−2rcosx+r2−π−πn=1n=116.^kCX½ny²)Z½n.+y²µÏ{Sn(x)}3[a,b]þÂñuS(x)§é∀ε>0,∀x∈[a,b],∃N(ε,x)∈Z§¦n>N(ε,x)§ÑAk|Sn(x)−S(x)|<ε§AOk|SN(ε,x)−S(x)|<εdSN(ε,x)(x)−S(x)3x:ëY§3x:mOx§¦|SN(ε,x)(y)−S(y)|,∀y∈Oxu´Oxx∈[a,b]¤[a,b]mCX(éà:a,bëYòÿ)âkCX½n§l¥ÀÑkmOx1,···,OxmÓCX[a,b]÷v|SN(ε,xi)(y)−S(y)|<ε,∀y∈Ox,i=1,2,···,miSmN=maxN(ε,xi)§Kn>N§é∀x∈[a,b]§d{Sn(x)}üN5ÚOx⊃[a,b]§73,ii6i6mi=1Ox§¦x∈Ox§k|Sn(x)−S(x)|6|SN(x)−S(x)|6|SN(ε,xi)(x)−S(x)|<εii={Sn(x)}3[a,b]þÂñuS(x).17.eSn(x)3c:ëY(n=1,2,3,···)§{Sn(c)}uѧK3?Ûm«m(c−δ,c)S(δ>0)§{Sn(x)}7ØÂñ.y²µ^y{.b3δ0>0§¦{Sn(x)}3(c−δ0,c)SÂñ++dÂñÜn§é∀ε>课后答案网0,∃N(ε)∈Z§¦n>N(ε)§é∀x∈(c−δ0,c)Ú∀p∈Z§ÑAεk|Sn+p(x)−Sn(x)|<(∗)¤á2ÏzSn(x)3c:ëY§KSn+p(x)−Sn(x)3c:ëYu´lim[Sn+p(x)−Sn(x)]=Sn+p(c)−Sn(c)x→c−0ε3(∗)ªüà-x→c−0§|Sn+p(c)−Sn(c)|6<ε2dêÜÂñn§{Sn(c)}www.hackshp.cnÂñ§®{Sn(c)}uÑgñbØ(§K3?Ûm«m(c−δ,c)S(δ>0)§{Sn(x)}7ØÂñ.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn184§2.?ê1.¦e?êÂñ«mµX∞(2x)n(1)n!n=1X∞ln(n+1)n+1(2)xn+1n=1"!n#nX∞n+1(3)xnn=1∞n2Xx(4)2nn=1X∞[3+(−1)n]nn(5)xnn=1X∞3n+(−2)nn(6)(x+1)nn=1)µn2(1)an=n!anÏR=lim=+∞§KÙÂñ(−∞,+∞).n→∞an+1X∞ln(n+1)X∞lnnlnnn+1n(2)x=x§an=n+1nnn=1n=2(y+1)lnyy+1lnyandulim=limlim=1§KR=lim=1§u´ÙÂñ«my→+∞yln(y+1)y→+∞yy→+∞ln(y+1)n→∞an+1(−1,1)X∞lnnnnx=−1§?ê(−1)xn!n=2课后答案网!()00lnx1−lnxlnxlnnÏ=x>3§<0§KüN~xx2xnlnnX∞lnnX∞lnnnnnnqlim=0§K?ê(−1)x4ÙZ[?ê§u´?ê(−1)xÂñn→∞nnnn=2n=2X∞lnnnx=1§?êxnwww.hackshp.cnn=2lnnX∞1X∞lnnnnÏlim=+∞§Kâ?ê"O{9?êuѧ?êxuÑn→∞nnnn=1n=2Kd?êÂñ[−1,1)."!n#n!n2!n2X∞n+1X∞11n(3)Ïx=1+x§Kan=1+nnnn=1n=1!p111qlimn|a.n|=e§KÙÂñ»R=§Âñ«m−,n→∞eee!n2!n!n2!n1X∞1111nnx=±§?ê(±1)1+§Kun=(±1)1+enenen=1−1dâ7{K§lim|un|=e26=0n→∞!n2!n!X∞1111nK?ê(±1)1+uѧu´?êÂñ−,.neeen=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn1851(4)an=2nrnp2n2n1dlim|an|x=|x|lim=|x|<1§ÙÂñ»R=1§Âñ«m(−1,1)n→∞n→∞2∞n2X(±1)|x|=1=x=±1§?êC2nn=1X∞n2X∞X∞n2(±1)1(±1)du?ê=Âñ§K?êýéÂñKÂñ2n2n2nn=1n=1n=1∞n2Xxl?êÂñ[−1,1].2nn=1nn[3+(−1)](5)an=rn!nnn[3+(−1)]111Ïlim=4§K?êÂñ»R=§Âñ«m−,n→∞n4441X∞[3+(−1)n]nX∞1X∞1x=§?êC=+4n·4n2k(2k+1)22k+1n=1k=1k=1X∞1é?ê(2k+1)22k+1k=11X∞(2k+3)22k+311Ïlim=<1§KâKO{§?êÂñk→∞14(2k+1)22k+1k=1(2k+1)22k+1X∞1X∞[3+(−1)n]nq?êuѧK?êuÑ2kn·4nk=1n=11X∞[3+(−1)n]nnÓ{§x=−§?ê(−1)uÑ4n·4nn=1!X∞[3+(−1)n]n11nK?êxÂñ−,.n44n=1nn3+(−2)(6)an=n!an1142Ïlim=§K?êÂñ»R=§Âñ«m−,−n→∞an+13课后答案网333X∞nn!nX∞nX∞ 2n43+(−2)1(−1)n3x=−§?êC(−1)=+3n3nn nn=1n=1n=1X∞23é?ênn=1 2n+1X∞ 2n/(n+1)2Ïlim3 =www.hackshp.cn<1§KâKO{§3Âñ2nn→∞/n3n3n=1X∞(−1)n4q?êÂñ§Kx=−§?êÂñ¶n3n=12Ó{§x=−§?êuÑ3"!X∞3n+(−2)n42nK?ê(x+1)Âñ−,−.n33n=12.¦?êÂñ»µ!X∞11n(1)1++···+x2nn=1X(2n)!n(2)x(n!)2)µ若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn18611(1)an=1++···+r2rnn1n11√nÏ1=n·61++···+6n·1→1(n→∞)pn2nKlimn|an|=1§u´ÙÂñ»R=1.n→∞(2n)!(2)an=(n!)2a(n+1)211nÏlim=lim=§u´ÙÂñ»R=.n→∞an+1n→∞2(n+1)(2n+1)44X∞3n+(−2)nXXnnn3.(x+1)?êanxÂñ»R§bnxÂñ»Q§?Øe?êÂñnn=1»µX2n(1)anxXn(2)(an+bn)xXn(3)anbnx)µrpp1211√(1)lim2n|an√n|=lim|an|==§KÙÂñ»R1=R.n→∞n→∞RR(2)Anp=an+bnppp√p√ppKkn|Annnnnnnn√n|=|an+bn|6|an|+|bn|62max(|an|,|bn|)=2·max(|an,bn|)=2max(|an|,|bn|)nÏlim2=1n→∞1p√ppppppK=limn|Annnnnnnn|6lim{2max(|an|,|bn|)}=lim{max(|an|,|bn|)}=maxlim|an|,lim|an|=R2n→∞!n→∞n→∞n→∞n→∞11max,RQ1l§R2>!=min(R,Q).11max,RQ(3)Bnp=anbnpp课后答案网pn|BnnnKkn|=|anbn|=|an|·|bn|1ppppp111u´=limn|Bnnnnn|=lim{|an|·|bn|6lim|an|·lim|bn|}=·=R3n→∞n→∞n→∞n→∞RQRQlR3>RQ.XXnn4.é¿©n§|an|6|bn|§@"?êanxÂñ»ØubnxÂñ».ppppy²µÏé¿©n§|awww.hackshp.cnnnnnn|6|bn|§K|an|6|bn|§u´lim|an|6lim|bn|XXn→∞n→∞nn?êanxÂñ»R§?êbnxÂñ»Qpp11K0Q¶n→∞n→∞limn|an|limn|bn|ppn→∞n→∞n|an0=limn|6lim|bn|§KR=∞,Q6∞§u´R>Q¶n→∞pn→∞pn|anlimn|6lim|bn|=∞§KR>0,Q=0§u´R>Qn→∞Xn→∞Xnnnþ§?êanxÂñ»ØubnxÂñ».5.y²?ê51Ú52.y²µ51.x(x0−R,x0+R)S?:§o±À00§½3õªQ(x)§¦||f(x)−Q(x)||=max|f(x)−Q(x)|0§kϕ(z)dz→(p→∞)πaz2y²µ(1)Ïϕ(x)3[a,b]þüNO¼ê§Kϕ(−t)3[−b,−a]þüN~¼êa=0,b<0§ϕ(−t)3[0,−b](−b>0)þüNO¼êZZZbsinpzbsinpz−bsinptéϕ(z)dzCþz=−t§Kϕ(z)=−ϕ(−t)dtazZaz0Zt−bsinptπbsinpzπKd)á4Ún§limϕ(−t)dt=ϕ(−0)=limϕ(z)dz=−ϕ(−0)Zp→∞0t2p→∞az21bsinpz1u´ϕ(z)dz→−ϕ(−0)(p→∞)πaz2ZZZbsinpz0sinpzbsinpz(2)Ïa<0,b>0§ϕ(x)3[a,b]þüNO¼ê§ϕ(z)dz=ϕ(z)dz+ϕ(z)dzZaZzaz0zasinpzπ0sinpzπâ(1)§limϕ(z)dz=−ϕ(−0)§Klimϕ(z)dz=ϕ(−0)p→∞0zZ2p→∞az2bsinpzπqd)á4Ún§limϕ(z)dt=ϕ(+0)p→∞0z2ZbsinpzπKlimϕ(z)dz=[ϕ(+0)+ϕ(−0)]p→∞az2Z1bsinpzϕ(−0)+ϕ(+0)u´ϕ(z)dz→(p→∞)πaz2课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn202§2.LpDC1.f(x)3(−∞,+∞)Sýéȧy²fb(ω)3(−∞,+∞)SëY.000000y²µé∀ω∈(−∞,+∞)§okA,A§¦ω∈[A,A]Z+∞Z+∞dufb(ω)=f(x)e−iωxdx6|f(x)|dx−∞−∞Z+∞−iωx000 öÂñعëþω§ùL²È©fb(ω)=f(x)edx3[A,A]þÂñ−∞000âÂñÈ©ëY5§fb(ω)3[A,A]þëY§l3:ω?ëYdω?¿5§fb(ω)3(−∞,+∞)SëY.2.f(x)3(−∞,+∞)Sýéȧy²limfb(ω)=0.ω→∞Z+∞εy²µdf(x)3(−∞,+∞)Sýéȧéu?ε>0§3A>0§¦k|f(x)|dx0§¦ω>δ§ðk|mk|<ω3k=1k=1u´éþãZ¤Àδ§ω>δZZZ+∞A+∞+∞f(x)sinωxdx6f(x)sinωxdx+f(x)sinωxdx6ε=limf(x)sinωxdx=0ω→∞00A0Ùg§f(x)3«m[0,A]¥k×:§{Bå§Øk×:0Zηεu´é?ε>0§3η>0§¦k|f(x)|dx<03ZAεqf(x)3[η,A]þÃa:§A^þã(J3䧦ω>δ§ðkf(x)sinωxdx<课后答案网η3ZZZZ+∞ηA+∞u´ω>δ§kf(x)sinωxdx6|f(x)|dx+f(x)sinωxdx+|f(x)|dx<εZ00ηA+∞=limf(x)sinωxdx=0ω→∞0Z+∞Ó{§f(x)3(−∞,+∞)Sýéȧþklimf(x)sinωxdx=0ω→∞www.hackshp.cnZ−∞+∞Ó{yf(x)3(−∞,+∞)Sýéȧlimf(x)cosωxdx=0ω→∞−∞u´limfb(x)=0.ω→∞3.¦e¼êLpDCµπEsinω0x,|x|<(1)f(x)=ω0π0,|x|>ω0π0,−∞0,∃N∈Z§n>N§kr(Mn,M0)<εpn→∞=(xn−x0)2+(yn−y0)2<εu´½k|xn−x0|6r(Mn,M0)<ε,|yn−y0|6r(Mn,M0)<ε=xn→x0,yn→y0(n→∞)⇐pÏ(|x22222n−x0|+|yn−y0|)>|xn−x0|+|yn−y0|=06|xn−x0|+|yn−y0|6|xn−x0|+|yn−y0|pqxn→x0,yn→y0(n→∞)§K|xn−x0|2+|yn−y0|2→0(n→∞)=(xn,yn)→(x0,y0)(n→∞)2.y²µe²¡þ:{Mn}Âñ§K§k4.0y²µlimMn=M0§bqklimMn=M0n→∞n→∞εε+0d½Â§é∀ε>0,∃N∈Z§n>N§kr(Mn,M0)<,r(Mn,M0)<2200dnØ ª§kr(M0,M0)6r(Mn,M0)+r(Mn,M0)<ε000qM0,M0½ü:§dε?¿5§r(M0,M0)=0=M0=M0.3.y²µeMn→M0(n→∞)§@o§?ÛfMn→M0.k+y²µÏMn→M0(n→∞)§Ké∀ε>0,∃N∈Z§n>N§kr(Mn,M0)<ε8K=N§Kék>K§knk>nK=nN>N§g,kr(Mn,M0)<ε=Mn→M0(k→∞).kk4.¦e:8ES:§:§>.:µ2(1)Ed÷vyx:(x,y)´E:¶÷vy=x:(x,y)´E>.:.222yyy222(2)÷v14:(x,y)´E:¶4www.hackshp.cn4422yy22÷vx+=1½x+=4:(x,y)´E>.:.442222(3)÷v01:(x,y)´E:¶:θ9÷22vx+y=1:(x,y)´E>.:.(4)dknê9ÃnêÈ5§²¡þ¤k:(x,y)Ñ´E>.:.5.y²µeM0´²¡:8Eà:§K3E¥3:Mn→M0(n→∞).1y²µ®M0´²¡:8Eà:§δn=§3O(M0,δ1)¥½3E:M16=M0¶3O(M0,δ2)¥½n3E:M2,M26=Mi(i6=0,1)1Xd?1e§:{Mn}(Mn6=Mi)(i=0,1,···,n−1)r(M0,Mn)0,∃N∈Z§n,m>N§kr(Mn,M0)<,r(Mm,M0)<22若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn205dålnØ ª§r(Mm,Mn)6r(Mn,M0)+r(Mm,M0)<ε⇐+:{Mn}÷vé∀ε>0,∃N∈Z§n,m>N§kr(Mn,Mm)<εò{Mn}©OÝKüI¶þ§ê{xn},{yn}Ï|xm−xn|0y61(2)½Â÷vØ ªy6x+1:8(3)½Â²¡x+y<022(4)½Â÷vØ ª2kπ6x+y6(2k+1)π(k=0,1,2,···):822222(5)½Â÷vØ ªr6x+y+z6R:82.¦e4µ22x+y(1)limx→0|x|+|y|y→022x+y(2)limpx→0x2+y2+1−1y→0221+x+y(3)limx→0x2+y2y→033sin(x+y)(4)limx→0x2+y2y→022−(x+y)(5)lim(x+y)ex→+∞y→+∞yln(x+e)(6)limpx→1x2+y2y→0课后答案网)µ22222x+y(|x|+|y|)x+y(1)Ï066=|x|+|y|lim(|x|+|y|)=0§Klim=0|x|+|y||x|+|y|x→0x→0|x|+|y|y→0y→022t√x+y(2)Ïlim√=lim(t+1+1)=2§Klimp=2t→+0t+1−1t→+0www.hackshp.cnx→0x2+y2+1−1y→0221+t1+x+y(3)Ïlim=+∞§Klim=+∞t→+0tx→0x2+y2y→0sin(x3+y3)|x3+y3||x|3+|y|3|x|3|y|3(4)Ï0666=+6|x|+|y|lim(|x|+|y|)=0x2+y2x2+y2x2+y2x2+y2x2+y2x→0y→033sin(x+y)Klim=0x→0x2+y2y→02tt(5)Ïlim=0,lim=0t→+∞ett→+∞et"#2(x+y)xy22−(x+y)Klim(x+y)e=lim−2·=0x→+∞x→+∞e−(x+y)exeyy→+∞y→+∞yln(x+e)(6)limp=ln2x→1x2+y2y→0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn2073.Áyelimf(x,y)=A3§x?ÛaC§4limf(x,y)=ϕ(x)3§Kg43§y→ay→bx→b uAµlimlimf(x,y)=limf(x,y)=Ax→ay→by→ax→b.22y²µÏ43§Ké∀ε>0,∃δ>0§|x−a|<δ,|y−b|<δ(x−a)+(y−b)6=0§ðk|f(x,y)−A|<εy30<|x−a|<δ¥½x§3þª¥-y→b§=|ϕ(x)−A|6ε§ùÒy²limϕ(x)=Ax→au´limlimf(x,y)=limϕ(x)=A=limf(x,y)x→ay→bx→ay→ax→b4.(1)ÁÞÑüg4Ø ~f¶(2)ÁÞÑkg43~f¶(3)ÁÞÑ43§g4Ø 3~f.)µx−y,x+y6=0(1)~µf(x,y)=x+y3:(0,0)g40,x+y=0x−ylimlimf(x,y)=lim=1,limlimf(x,y)=lim=−1x→0y→0x→0xy→0x→0y→0yKlimlimf(x,y)6=limlimf(x,y).x→0y→0y→0x→01xsin+yx(2)~µf(x,y)=3:(0,0)g4x+yylimlimf(x,y)=lim=1y→0x→0y→0y1xsinx1limlimf(x,y)=lim=limsinØ3.x→0y→0x→0xx→0x1xsin,y6=0(3)~µf(x,y)=y3:(0,0)g4Ú40,y=01Ï06|f(x,y)|=xsin6|x|§Klimf(x,y)=0=Ù43yx→0y→01limlimf(x,y)=0§y课后答案网→0§xsin4Ø3§=limlimf(x,y)Ø3.y→0x→0yx→0y→05.?Øe¼ê3:(0,0)g4Ú4µx2y2(1)f(x,y)=x2y2+(x−y)211(2)f(x,y)=(x+y)·sin·sinwww.hackshp.cnxy)µ(1)limlimf(x,y)=0,limlimf(x,y)=0x→0y→0y→0x→022xkeUy=kx→04§Kklimf(x,y)=limy=kxx→0x2k2+(1−k)2x→0AO§©Ok6=19k=1§BØÓ4091§Ïdlimf(x,y)Ø3.x→0y→0(2)Ï06|f(x,y)|6|x+y|6|x|+|y|§Klimf(x,y)=0=Ù43x→0y→0!11111qlim(x+y)·sin·sinØ3x6=(k=±1,±2,···)§lim(x+y)·sin·sinØy→0xykπx→0xy!13y6=(k=±1,±2,···)kπ=limlimf(x,y)9limlimf(x,y)ÑØ3.x→0y→0y→0x→0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn2086.?Øe¼êëYµ1(1)u=px2+y222(2)u=ln(1−x−y)1(3)u=sinxsiny1(4)u=ln(x−1)a+(y−b)2+(z−c)2)µ1(1)¼êu=p3:(0,0)ý§:(0,0)d¼êØëY:§Ød:þëY¶x2+y22222(2)üS:§=÷vx+y<1:¼êu=ln(1−x−y)ëY:¶(3)ëYx6=mπ,y6=nπ(m,n=0,±1,±2,···).(4)Ø:(a.b.c)þëY.7.y²¼ê2xy22,x+y6=0f(x,y)=x2+y2220,x+y=0©OéuzCþxÚy´ëY§"uCþëY¼ê.2axy²µk½y=a6=0§Kx¼êg(x)=f(x,a)=(−∞0§|x−x0|<δ1§εk|f(x,y0)−f(x0,y0)|<2qÏf(x,y)3GSéy÷voÊF[^§Ké?Û(x,y)∈G,(x,y0)∈G§k|f(x,y)−f(x,y0)|6L|y−y0|若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn209εε-L|y−y0|<§K|y−y0|<2!2Lεδ=minδ1,§|x−x0|<δ,|y−y0|<䧽k2L|f(x,y)−f(x0,y0)|6|f(x,y)−f(x,y0)|+|f(x,y0)−f(x0,y0)|<ε=d¼ê3GSëY.课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn2101oÙêÚ ©§1.êÚ ©Vg1.¦e¼êêµ222(1)z=xln(x+y)xy(2)u=ex(3)z=xy+yy(4)u=arctanx222(5)u=x+y+z+2xy+2yz+2zxϕ−θ(6)u=ecos(θ+ϕ))µ"#22x2xy22(1)zx=2xln(x+y)+,zy=.x2+y2x2+y2xyxy(2)ux=ye,uy=xe.21x(y−1)(3)zx=y+,zy=.yy2yx(4)ux=−,uy=.x2+y2x2+y2(5)ux=2(x+y+z),uy=2(x+y+z),uz=2(x+y+z).ϕ−θϕ−θ(6)uϕ=e[cos(θ+ϕ)−sin(θ+ϕ)],uθ=−e[sin(θ+ϕ)+cos(θ+ϕ)].222.f(x,y)=xy−2y§¦fx(x,y),fy(x,y),fx(2,3),fy(0,0),fy(x,y)x=y.y=x222)µfx(x,y)=2xy,fy(x,y)=2xy−2,fx(2,3)=36,fy(0,0)=−2,fy(x,y)x=y=2xy−2y=x√√∂z∂z13.z=ln(x+y)§y²x+y=.∂x∂y2√√∂z1∂z1y²µÏz=ln(x+y)§K=√√√,=√√√∂x课后答案网2x(x+y)∂y2y(x+y)∂z∂z1u´x+y=.∂x∂y24.¦e¼ê3½:(x0,y0) ©µ4422(1)u=x+y−4xy,(0,0),(1,1)xwww.hackshp.cn(2)u=p,(1,0),(0,1)x2+y2!ππ(3)u=xsin(x+y),(0,0),,442(4)u=ln(x+y),(0,1),(1,1))µ2222(1)Ïdu=4x(x−2y)dx+4y(y−2x)dy§K3(0,0):du=0¶3(1,1):du=−4dx−4dy.2yxy(2)Ïdu=dx−dy§K33(x2+y2)2(x2+y2)23(1,0):du=0¶3(0,1):du=dx.(3)Ïdu=[sin(x+y)+xcos(x+y)]dx+xcos(x+y)dy§K!ππ3(0,0):du=0¶3,:du=dx.44若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn211dx2y(4)Ïdu=+dy§Kx+y2x+y2dx3(0,1):du=dx+2dy¶3(1,1):du=+dy.25.¦e¼ê ©µ22(1)u=sin(x+y)mn(2)u=x·yxy(3)u=ey(4)u=xp(5)u=x2+y2+z2222(6)u=ln(x+y+z))µ22(1)du=2cos(x+y)(xdx+ydy)m−1n−1(2)du=xy(mydx+nxdy)xy(3)du=e(ydx+xdy)y−1(4)du=x(ydx+xlnxdy)xdx+ydy+zdz(5)du=px2+y2+z22(xdx+ydy+zdz)(6)du=x2+y2+z2p6.y²µf(x,y)=|xy|3(0,0)ëY§fx(0,0),fy(0,0)3§3(0,0):Ø.py²µdlim|xy|=0§limf(x,y)=0=f(0,0)§Kf(x,y)3(0,0):ëYx→0x→0y→0y→0f(∆x,0)−f(0,0)f(0,∆y)−f(0,0)Ïfx(0,0)=lim=0,fy(0,0)=lim=0∆x→0∆x∆y→0∆yKfx(0,0),fyp(0,0)3f(x,y)=|xy|3(0,0):Ø.e§Kk∆f=fx(0,0)∆x+fy(0,0)∆y+o(ρ)=∆f=o(ρ)p课后答案网∆f|∆x∆y|1Ä:P(x,y)÷y=xªu0§k=p=√6→0(ρ→0)gñ§u´bؤá§ρ∆x2+∆y22Kf(x,y)3(0,0):Ø.xy22p,x+y6=07.y²µf(x,y)=x2+y23(0,0):¥ëY§fx(x,y),fy(x,y)k.§3(0,0):220,xwww.hackshp.cn+y=0Ø.xy22y²µdup´Ð ¼ê§3Ù½ÂS7ëY§Kf(x,y)3x+y6=0ëYx2+y2pxy|xy|x2+y2q00)∂z∂zpu´x+y=z=x2+y2.∂x∂y9.ϕψ´?¿ ¼ê§y²µ!!yyz=xϕ+ψxx若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn216222∂z∂z∂z22÷vx+2xy+y=0∂x2∂x∂y!∂y2!!!!∂zyyyyy∂zy1y0000y²µÏ=ϕ−ϕ−ψ,=ϕ+ψ∂xxxxx2x∂yxxx22222∂zy2yy∂zy1y∂z1100000000000000K=ϕ+ψ+ψ,=−ϕ−ψ−ψ,=ϕ+ψ∂x2x3x3x4∂x∂yx2x2x3∂y2xx2222∂z∂z∂z22u´x+2xy+y=0∂x2∂x∂y∂y210.u=ϕ(x+at)+ψ(x−at)§Ù¥ϕ,ψ´?¿g¼ê§¦y22∂u∂u2=a.∂t2∂x2y²µÏu=ϕ(x+at)+ψ(x−at)§ϕ,ψ´?¿g¼ê22∂u∂u∂u∂u0000200000000K=a(ϕ−ψ),=ϕ+ψ§u´=a(ϕ+ψ),=ϕ+ψ∂t∂x∂t2∂x222∂u∂u2l=a.∂t2∂x2课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn217§3.d§(|)¤(½¼ê¦{1.¦de§¤(½¼êz=f(x,y) Ú êµz(1)x+y+z=e(2)xyz=x+y+z)µz1ez(1)ü>"ux¦§1+zx=zxe§Kzx=§u´zx2=ez−1(1−ez)3zz1eeÓ{§zy=,zy2=,zxy=zyx=ez−1(1−ez)3(1−ez)3yz−1(2)ü>"ux¦§yz+xyzx=1+zx(∗)§Kzx=1−xy2yzx2y(yz−1)ò(∗)ªü>"ux¦§2yzx+xyzx2=zx2§Kzx2==1−xy(xy−1)2xz−12x(xz−1)2zÓ{§zy=,zy2=,zxy=zyx=1−xy(xy−1)2(xy−1)22.¦de§¤(½¼ê ©½êµ∂z∂z(1)f(x+y,y+z,z+x)=0§¦,¶∂x∂y(2)z=f(xz,z−y)§¦dz¶∂z∂z(3)F(x−y,y−z,z−x)=0§¦,¶∂x∂y2∂z∂z∂z(4)F(x,x+y,x+y+z)=0§¦,,.∂x∂y∂x2)µf1+f3(1)ü>"ux¦§z=z(x,y)§f1+f2zx+f3(zx+1)=0§Kzx=−f2+f3f1+f2Ó{§zy=−f2+f3zf1dx−f2dy(2)ü੧dz=(xdz+zdx)f1+(dz−dy)f2§Kdz=课后答案网1−xf1−f2F1−F3(3)ü>"ux¦§z=z(x,y)§F1−F2zx+F3(zx−1)=0§Kzx=F2−F3F2−F1Ó{§zy=F2−F3F1+F2+F3(4)ü>"ux¦§z=z(x,y)§F1+F2+F3(1+zx)=0(∗)§Kzx=−www.hackshp.cnF33(∗)ªü>2"ux¦§F11+F12+F13(1+zx)+F21+F22+F23(1+zx)+zx2F3+(1+zx)[F13+F23+F33(1+zx)]=0122Kzx2=−3[F3(F11+2F12+F22)−2F3(F1+F2)(F13+F23)+F33(F1+F2)]F3F2+F3Ó{§zy=−F303.d§z=x+y·ϕ(z)(½¼êz=z(x,y)§1−yϕ(z)6=0§y²∂z∂z=ϕ(z)·∂y∂x0y²µ§ü੧z=z(x,y)§dz=dx+ϕ(z)dy+yϕ(z)dzdx+ϕ(z)dy0q1−yϕ(z)6=0§Kdz=1−yϕ0(z)∂zϕ(z)∂z1∂z∂zu´=,=§l=ϕ(z)·∂y1−yϕ0(z)∂x1−yϕ0(z)∂y∂x若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn218∂z∂z2224.y²d§ax+by+cz=Φ(x+y+z)¤½Â¼êz=z(x,y)÷v§(cy−bz)+(az−cx)=∂x∂ybx−ay§Ù¥Φ(u)´u¼ê§a,b,c~ê.y²µ§ü੧z=z(x,y)§Φ(u)´u¼ê0Kadx+bdy+cdz=2(xdx+ydy+zdz)Φ00∂z2xΦ−a∂z2yΦ−bu´=,=∂xc−2zΦ0∂yc−2zΦ0∂z∂zl(cy−bz)+(az−cx)=bx−ay∂x∂y∂z∂z5.ϕ?¿¼ê§y²d§ϕ(cx−az,cy−bz)=0¤½Â¼êz=z(x,y)÷va+b=c.∂x∂yy²µé§üà©O"ux,y¦§z=z(x,y)§cϕ1−aϕ1zx−bϕ2zx=0,−aϕ1zy+cϕ2−bϕ2zy=0∂zcϕ1∂zcϕ2u´=,=∂xaϕ1+bϕ2∂yaϕ1+bϕ2∂z∂zla+b=c.∂x∂y∂z∂z−1−16.y²d§F(x+zy,y+zx)=0¤(½¼êz=z(x,y)÷vx+y=z−xy.∂x∂yy²µé§üà©O"ux,y¦§z=z(x,y)§!!!!zxzxzzyzzyF11++F2−=0,F1−+F21+=0yxx2yy2x22∂zyzF2−xyF1∂zxzF1−xyF2u´=,=∂xx(xF1+yF2)∂yy(xF1+yF2)∂z∂zlx+y=z−xy.∂x∂y7.¦e§|¤(½¼êê½ê½ ©µ2x+y+z=0,dydzdy(1)¦,,¶x·y·z=1,dxdxdx2x+y=u+v,(2)xsinu¦du,dv¶=,ysinv2xu+yv=0,∂u∂u课后答案网∂v∂v∂u(3)¦,,,,¶yu+xv=1,∂x∂y∂x∂y∂x∂yx=cosθcosϕ,∂z∂z(4)y=cosθsinϕ,¦,¶∂x∂yz=sinθ,u=f(u,x,v+y),∂u∂v(5)2¦www.hackshp.cn,.v=g(u−x,u·y),∂x∂x)µdydz1++=0(1)éx¦§dxdx(∗)dydzyz+xz+xy=0dxdxdyy(z−x)dzz(x−y)éá¦)§=,=dxx(y−z)dxx(y−z)22dydz+=0(∗)ª2éx¦§dx2dx222dydzdydydzdydzdydzdzz+y+z+x·+xz+y+x·+xy=0dxdxdxdxdxdx2dxdxdxdx2d2y2zdy+2ydz+2xdy·dzdxdxdxdxéá§=dx2x(y−z)2222dydzdyyz[(x−y)+(x−z)+(y−z)]ò,§=dxdxdx2x2(z−y)3若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn219u+v=x+ydu+dv=dx+dy(2)òªU©§ysinu=xsinvsinudy+ycosudu=sinvdx+xcosvdv1Kdu=[(sinv+xcosv)dx−(sinu−xcosv)dy]xcosv+ycosu1dv=[−(sinv−ycosu)dx+(sinu+ycosu)dy]xcosv+ycosuxdu+ydv=−udx−vdy(3)©§ydu+xdv=−vdx−udy11u´du=[(yv−xu)dx+(yu−xv)dy],dv=[(yu−xv)dx+(yv−xu)dy]x2−y2x2−y2∂uyv−xu∂uyu−xv∂vyu−xv∂uyv−xuK=,=,=,=∂xx2−y2∂yx2−y2∂xx2−y2∂xx2−y2222∂u(yux−v−xvx)(x−y)−2x(yu−xv)u´=∂x∂y(x2−y2)2222∂u∂v∂u2(xv+yv−2xyu)ò,§=∂x∂x∂x∂y(x2−y2)2∂θ∂ϕ1=−sinθ·cosϕ−cosθ·sinϕ(4)dx,yéx¦ê§∂x∂x∂θ∂ϕ0=−sinθ·sinϕ+cosθ·cosϕ∂x∂x∂θcosϕ∂ϕsinϕ∂z∂z∂θxK=−,=−§u´=·=−cotθcosϕ=−∂xsinθ∂xcosθ∂x∂θ∂xz∂zyÓn§=−∂yz∂u∂u∂v=f1+f2+f3∂x∂x!∂x(5)éx¦§∂v∂u∂v=g1−1+2vyg2∂x∂x∂x∂uf2(1−2vyg2)−g1f3∂vg1(f1+f2−1)K=,=.∂x(f1−1)(2vyg2−1)−g1f3∂x(f1−1)(2vyg2−1)−g1f3∂z∂z22338.§x=u+v,y=u+v,z=u+v½Âzx,y¼ê§¦,.∂x∂yx2222)µÏx−y=2uv§Kz=(u+v)(u−uv+v)=(3y−x)课后答案网2∂z3∂z32u´=(y−x),=x.∂x2∂y29.x=rcosθ,y=rsinθ§C§dx22=y+kx(x+y)dtdy22www.hackshp.cn=−x+ky(x+y)dt4I§.)µd§§x,y´t¼ê§l4ICr,θ´t¼ê§x=rcosθ,y=rsinθdxdrdθ=cosθ−rsinθüàét¦§dtdtdtdydrdθ=sinθ+rcosθdtdtdtdrdθ2dxdycosθ−rsinθ=rsinθ+krcosθ·ròx,y,,§|§dtdtdtdtdrdθ2sinθ+rcosθ=−rcosθ+krsinθ·rdtdtdrdθ3u´=kr,=−1.dtdt22∂z∂zuu10.x=ecosθ,y=esinθ§C§+=0.∂x2∂y2yuu22)µÏx=ecosθ,y=esinθ§Ku=ln(x+y),θ=arctanx若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn220∂ux∂uy∂θy∂θx∂u∂θ∂u∂θK=,=;=−,=§u´=,=−∂xx2+y2∂yx2+y2∂xx2+y2∂yx2+y2∂x∂y∂y∂x∂z∂z∂u∂z∂θ∂z∂z∂u∂z∂θq=·+·,=·+·∂x∂u∂x∂θ∂x∂y∂u∂y∂θ∂y!2!2222222∂z∂z∂u∂z∂u∂θ∂z∂θ∂z∂u∂z∂θK=+2··++·+·∂x2∂u2∂x∂θ∂u∂x∂x∂θ2∂x∂u∂x2∂θ∂x2!2!2222222∂z∂z∂u∂z∂u∂θ∂z∂θ∂z∂u∂z∂θ=+2··++·+·∂y2∂u2∂y∂θ∂u∂y∂y∂θ2∂y∂u∂y2∂θ∂y2!!!22∂u∂∂θ∂∂θ∂∂u∂uq===−=−∂x2∂x∂y∂y∂x∂y∂y∂y222∂θ∂θÓ{§=−∂x2∂y22222∂θ∂θ∂u∂uK+=+=0∂x2∂y2∂x2∂y2!2!2!2!2∂u∂u∂θ∂θ∂u∂θ∂u∂θq+=+,·=−·∂x∂y∂x∂y∂x∂x∂y∂y!2222∂z∂z∂z∂z−2uK+=e+=0∂x2∂y2∂u2∂θ222∂z∂z=+=0.∂u2∂θ222∂f∂f11.x=rcosθ,y=rsinθ§Kf(x,y)=Φ(r,θ)§^Φ"ur,θê5L«+.∂x2∂y2)µòf(x,y)=Φ(r,θ)"ur,覧∂f∂x∂f∂y∂Φ∂f∂f∂Φ·+·==cosθ+sinθ=∂x∂r∂y∂r∂r∂x∂y∂r∂f∂x∂f∂y∂Φ∂f∂f∂Φ·+·==−rsinθ+rcosθ=∂x∂θ∂y∂θ∂θ∂x∂y∂θ2222∂Φ∂f∂f∂f22K=cosθ+sin2θ+sinθ∂r2∂x2∂x∂y∂y2!2222∂Φ∂f∂f∂f∂f∂f222=rsinθ−sin2θ+cosθ−rcosθ−rsinθ∂θ2∂x2∂x∂y∂y2∂x∂y2222∂Φ1∂Φ∂f∂课后答案网f1∂Φu´+=+−∂r2r2∂θ2∂x2∂y2r∂r2222∂f∂f∂Φ1∂Φ1∂Φl+=++∂x2∂y2∂r2r2∂θ2r∂r222∂z∂z∂zξη2212.x=e,y=e§C§ax+2bxy+cy=0(a,b,c~ê).∂x2www.hackshp.cn∂x∂y∂y2dξ1dη1ξη)µÏx=e,y=e§Kξ=lnx,η=lny§u´=,=dxxdyy∂z1∂z∂z1∂zK=,=∂xx∂ξ∂yy∂η!!222222∂z1∂z∂z∂z1∂z∂z∂z1∂zu´=−,=−,=∂x2x2∂ξ2∂ξ∂y2y2∂η2∂η∂x∂yxy∂ξ∂η!!222∂z∂z∂z∂z∂z§z{n§a−+2b+c−=0.∂ξ2∂ξ∂ξ∂η∂η2∂η∂z∂z2213.ξ=x,η=x+y§C§y−x=0.∂x∂y)µd§z´x,y¼ê§ξ,ηq´x,y¼ê§lzw¤´ÏL¥mCþξ,η"ux,yEܼê∂z∂z∂z∂z∂zu´=+2x,=2y∂x∂ξ∂η∂y∂η∂z∂z∂zÏy−x=y∂x∂y∂ξ若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn221∂z∂z∂zÏy6≡0§Kdy−x=0§=0.∂x∂y∂ξ∂u∂u∂u14.ξ=x,η=y−x,ζ=z−x§C§++=0.∂x∂y∂z)µd§u´x,y,z¼ê§ξ,η,ζq´x,y,z¼ê§luw¤´ÏL¥mCþξ,η,ζ"ux,y.zEܼê∂u∂u∂u∂u∂u∂u∂u∂uu´=−−,=,=∂x∂ξ∂η∂ζ∂y∂η∂z∂ζ∂u∂u∂u∂zKd++=0§=0∂x∂y∂z∂ξ222∂u∂u∂u15.5Cξ=x+λ1y,η=x+λ2y§y3r§A+2B+C=0(A,B,C~ê§∂x2∂x∂y∂y22∂u22AC−B<0)C=0§y²λ1,λ2§Cλ+2Bλ+A=0üÉ¢.∂ξ∂ηy²µd§u´x,y¼ê§Ï±ruÀ±ξ,η¥mCþ"ux,yEܼê§u´∂u∂u∂u∂u∂u∂u=+,=λ1+λ2∂x∂ξ∂η∂y∂ξ∂η22222222∂u∂u∂u∂u∂u∂u∂u∂u22∂x2=∂ξ2+2∂ξ∂η+∂η2,∂y2=λ1∂ξ2+2λ1λ2∂ξ∂η+λ2∂η2,2222∂u∂u∂u∂u=λ1+(λ1+λ2)+λ2∂x∂y∂ξ2∂ξ∂η∂η2222222∂u∂u∂u∂u∂u∂u2ÏA+2B+C=(A+2Bλ1+Cλ1)+2[A+B(λ1+λ2)+Cλ1λ2]+(A+∂x2∂x∂y∂y2∂ξ2∂ξ∂η∂η222Bλ2+Cλ2)=02∂2u∂2u∂2u∂2uA+2Bλ1+Cλ1=02¦A+2B+C=0C=0§7LA+2Bλ2+Cλ2=0∂x2∂x∂y∂y2∂ξ∂ηA+B(λ1+λ2)+Cλ1λ26=02dcü§λ1,λ2´§Cλ+2Bλ+A=02d1n§λ16=λ2§Kλ1,λ2´Cλ+2Bλ+A=0üÉ¢2BA22qÏλ1+λ2=−,λ1λ1=§KA+B(λ1+λ2)+Cλ1λ2=(AC−B)6=0CCC2222∂u∂u∂u∂uu´§A+2B+C=035Cξ=x+λ1y,η=x+λ2ye(¢C=0§∂x2∂x∂y∂y2∂ξ∂η2λ1,λ2§Cλ+2Bλ+A=0üÉ¢.!22∂w∂课后答案网w∂ϕ∂ψ∂ϕ∂ψ16.y².Ê.d§∆w≡+=03Czx=ϕ(u,v),y=ψ(u,v)§÷v=,=−e∂x2∂y2∂u∂v∂v∂u/G±ØC.y²µl§ω´x,y¼ê§x,y´u,v¼ê§Kw´±x,y¥mCþu,v¼ê∂w∂w∂ϕ∂w∂ψ∂w∂w∂ϕ∂w∂ψu´=·+·,=·+·∂u∂x∂u∂y∂u∂v∂x∂v∂y∂v!2!2222222∂w∂w∂ϕ∂w∂ϕwww.hackshp.cn∂ψ∂w∂ψ∂w∂ϕ∂w∂ψ=+2··++·+·∂u2∂x2∂u∂x∂y∂u∂u∂y2∂u∂x∂u2∂y∂u2!2!2222222∂w∂w∂ϕ∂w∂ϕ∂ψ∂w∂ψ∂w∂ϕ∂w∂ψ=+2··++·+·∂v2∂x2∂v∂x∂y∂v∂v∂y2∂v∂x∂v2∂y∂v222222222∂ϕ∂ψ∂ϕ∂ψ∂ϕ∂ψ∂ϕ∂ψ∂ψ∂ϕ∂ψ∂ϕ5¿òz^=,=−§K=,=−,=−,=∂u∂v∂v∂u∂u2∂u∂v∂v2∂v∂u∂u2∂u∂v∂v2∂v∂u22∂w∂wò,§¿òþ㪧∂u2∂v2!"!!#222222∂w∂w∂w∂w∂ϕ∂ϕ+=++∂u2∂v2∂x2∂y2∂u∂v2222∂w∂w∂w∂wÏ+=0§K+=0∂x2∂y2∂u2∂v222∂w∂wùL².Ê.d§∆w≡+=03Czx=ϕ(u,v),y=ψ(u,v)e/G±ØC.∂x2∂y222∂u∂u217.ξ=x−at,η=x+at§C§=a.∂t2∂x2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn222)µd§u´t,x¼ê§ξ,η´t,x¼ê§òuÀ±ξ,η¥mCþ"ut,x¼ê∂u∂u∂u∂u∂u∂uK=−a+a,=+∂t∂ξ∂η∂x∂ξ∂η22222222∂u∂u∂u∂u∂u∂u∂u∂u222=a−2a+a,=++∂t2∂ξ2∂ξ∂η∂η2∂x2∂ξ2∂ξ∂η∂η2222∂u∂u∂u22u´d=a§4a=0∂t2∂x2∂ξ∂η2∂uqa6≡0§K=0.∂ξ∂η18.gCêÚÏCêC§u,v#gCê§w=w(u,v)#ÏCêµ1122(1)u=x+y,v=+,w=lnz−(x+y)§C§xy∂z∂zy−x=(y−x)·z∂x∂yyz(2)u=x+y,v=,w=§C§xx222∂z∂z∂z−2+=0∂x2∂x∂y∂y2uu(3)x=u,y=,z=§C§1+uv1+u·w∂z∂z222x+y=z∂x∂yx(4)u=,v=x,w=xz−y§C§y2∂z∂z2y+2=∂y2∂yx)µ111(1)d®§du=2xdx+2ydy,dv=−dx−dy,dw=dz−dx−dy课后答案网x2y2z∂w∂w,¡§dw=du+dv∂u∂v!1∂w∂w11Kdz−dx−dy=(2xdx+2ydy)+−dx−dyz∂u∂vx2y2!!∂wz∂w∂wz∂wn§dz=2xz−·+zdx+2yz−·+zdy∂ux2www.hackshp.cn∂v∂uy2∂v!∂z∂zxy∂wòþª¤(½,§z−=0∂x∂yy2x2∂v!xy∂wqz−6≡0§K=0.y2x2∂vy1z1(2)d®§du=dx+dy,dv=−dx+dy,dw=−dx+dzx2xx2x∂w∂w,¡§dw=du+dv∂u∂v!z1∂w∂wy1K−dx+dz=(dx+dy)+−dx+dyx2x∂u∂vx2x!!∂wy∂wz∂w∂wn§dz=x−·+dx+x+dy∂ux∂vx∂u∂v∂z∂wy∂wz∂z∂w∂wK=x−·+,=x+∂x∂ux∂vx∂y∂u∂v若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn223∂z∂z∂w-R=−=w−(1+v)∂x∂y∂v!!!222∂z∂z∂z∂∂z∂z∂∂z∂z∂R∂R∂R∂u∂uK−2+=−−−=−=−+∂x2∂x∂y∂y2∂x∂x∂y∂y∂x∂y∂x∂y∂u∂x∂y!"#!22∂R∂v∂v∂∂wy1(1+v)∂w−=w−(1+v)−−==0∂v∂x∂y∂v∂vx2xx∂v222(1+v)∂wÏ6≡0§K§C=0.x∂v2uu1111(3)Ïx=u,y=,z=§Ku=x,v=−,w=−1+uv1+u·wyxzx1111u´du=dx,dv=dx−dy,dw=dx−dzx2y2x2z2∂w∂w,¡§dw=du+dv∂u∂v!11∂w∂w11Kdx−dz=dx+dx−dyx2z2∂u∂vx2y2!21∂w1∂wz∂w2n§dz=z−−·dx+·dyx2∂ux2∂vy2∂v∂z∂z∂w22òþª¤(½,§xz=0∂x∂y∂u∂wqxz6≡0§K=0.∂ux∂w∂z∂w∂w∂u∂w∂v∂ux∂v(4)Ïu=,v=x,w=xz−y§K=x−1,=·+·,=−,=0y∂y∂y∂y∂u∂y∂v∂y∂yy2∂y∂wx∂w∂z11∂wu´=−§K=−∂yy2∂u∂yxy2∂u22∂z∂z2x∂w2u´y+2=+=∂y2∂yxy3∂u2x2x∂wq6≡0§K§C=0.y3∂u2课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn224§4.m{²¡1.¦e3¤«:?{²¡µπ22(1)x=asint,y=bsint·cost,z=ccost§3t=:?¶4222(2)x+y+z=6,x+y+z=0§3:(1,−2,1).)µabc000(1)x0=,y0=,z0=,x(t0)=a,y(t0)=0,z(t0)=−c222acx−2z−2xzπ=+=1K3t=:?§a−c=ac4bby=y=!!22ac122{²¡§ax−−cz−=0=ax−cz=(a−c).222D(F1,F2)2y2zD(F1,F2)2z2x(2)ÏD(y,z)=11=−6,D(z,x)=11=0,(1,−2,1)(1,−2,1)(1,−2,1)(1,−2,1)D(F1,F2)2x2y==6D(x,y)11(1,−2,1)(1,−2,1)x+z−2=0K3:(1,−2,1)§y=−2{²¡§x−z=0.232.3x=t,y=t,z=tþ¦Ñ:§¦d:²1u²¡x+2y+z=4.230002)µ¤¦:(t0,t0,t0)§Kx(t0)=1,y(t0)=2t0,z(t0)=3t02u´¥þv={1,2t0,3t0}12q²¡{¥þn={1,2,1}§KâK¿§Akv·n=1+4t0+3t0=0§u´t0=−1,t0=−!3111K¤¦:(−1,1,−1),−,,−.3927ttt2223.y²x=aecost,y=aesint,z=aeI¡x+y=z1¤Ó.22222t222t222t2y²µòx,y,zx+y=z§aecost+aesint=ae=z§KA3¡þ222Ix+y=zº:3:§LIþ?:P课后答案网(x0,y0,z0)1L:K1¥þv1={x0,y0,z0}q3:Pþvt0t0t02={ae(cost0−sint0),ae(sint0+cost0),ae}={x0−y0,x0+y0,z0}222x0+y0=z0√v1·v226Kcos(v1,v2)==√=§ùþ:(x,y,z)vk"X|v1||v2|63ÏI¡1¤Ó.www.hackshp.cn4.¦e3¤«:{uµ234(1)x=t,y=t,z=t§3t=1:þ¶2(2)xyz=1,y=x§3:(1,1,1).)µ000(1)Ïx(t0)=2,y(t0)=3,z(t0)=4§Kþ{2,3,4}2√3√4√u´{uµcosα=±29,cosβ=±29,cosγ=±29.292929D(F1,F2)xzxyD(F1,F2)xyyz(2)ÏD(y,z)=−2y0=2,D(z,x)=01=1,(1,1,1)(1,1,1)(1,1,1)(1,1,1)D(F1,F2)yzxz==−3§Kþ{2,1,−3}D(x,y)1−2y(1,1,1)√(1,1,1)√14143√u´{uµcosα=±,cosβ=±,cosγ=±14.71414若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn225§5.¡²¡{1.¦e¡3¤«:²¡9{§µ(1)x=asinϕcosθ,y=asinϕsinθ,z=acosϕ§3(θ0,ϕ0)¶xy(2)ez+ez=4§3:(ln2,ln2,1)¶22(3)z=2x+4y§3:(2,1,12)¶222(4)ax+by+cz+d=0§3:(x0,y0,z0).)µD(y,z)asinϕcosθacosϕsinθ2(1)Ï=0−asinϕ=−asinϕ0cosθ0,D(θ,ϕ)(θ0,ϕ0)(θ0,ϕ0)D(z,x)D(x,y)222=−asinϕ0sinθ0,=−asinϕ0cosϕ0D(θ,ϕ)D(θ,ϕ)(θ0,ϕ0)(θ0,ϕ0)K²¡§sinϕ0cosθ0x+sinϕ0sinθ0y+cosϕ0z=ax−asinϕ0cosθ0y−asinϕ0sinθ0z−acosϕ0{§==.sinϕ0cosθ0sinϕ0sinθ0cosϕ0(2)Ï3(ln2,ln2,1):fx=2,fy=2,fz=−ln16x−ln2y−ln2z−1K²¡§x+y−2ln2·z=0¶{§==.11−2ln2(3)Ïzx(2,1)=8,zy(2,1)=8x−2y−1z−12K²¡§8x+8y−z=12¶{§==.88−1(4)Ï3(x0,y0,z0):fx=2ax0,fy=2by0,fz=2cz0x−x0y−y0z−z0K²¡§ax0x+by0y+cz0z+d=0¶{§==.ax0by0cz02.3¡z=xyþ¦:§¦ù:{Ru²¡x+3y+z+9=0§¿Ñd{§.)µL¡þ?:M0(x0,y0,z0)n1={y0,x0,−1}§{þn2={1,3,1}−y−x1¦{Ruþ㲡§Kn1kn2===131x+3y+1z−3u´¤¦:(−3,−1,3)§K{§==.131√√√√3.y²¡x+y+z=a,(a>课后答案网0)þ?Û:²¡3I¶þ åÚ ua.y²µ3¡þ?:P0(x0,y0,z0)111K¡3T:²¡§√(x−x0)+√(y−y0)+√(z−z0)=02x02y02z0√√√=y0z0(x−x0)+x0z0(y−y0)+x0y0(z−z0)=0√√√√√√√u´²¡3I¶þ å©ax0,ay0,az0§ÙÚa(x0+y0+z0)=a.2224.¦ü¡x+y=a,bz=xy.www.hackshp.cn)µü¡?:M0(x0,y0,z0)dü¡3M0:{þn1={2x0,2y0,0},n2={y0,x0,−b}n1·n22bz0u´ϕ÷vcosϕ==√.|n1||n2||a|a2+b2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn226§6.êÚFÝ221.¦u=x−xy+y3(1,1)?÷l=(cosα,sinα)ê.¿?Ú¦µ(1)3=þÙêk¶(2)3=þÙêk¶(3)3=þÙê0¶(4)¦uFÝ.)µÏux=2x−y,uy=−x+2y§Kux(1,1)=1,uy(1,1)=1!∂u∂u√πq=ux(1,1)cosα+uy(1,1)sinα§K=cosα+sinα=2sinα+∂l∂l4u´√√!π22√(1)α=§3l=,þÙêk2¶422√√!322√(2)α=−π§3l=−,−þÙêk−2¶422√√!√√!π32222(3)α=−,π§3l=,−½l=−,þÙê0¶442222(4)gradu=ux(1,1)i+uy(1,1)j=i+j.2.¦u=xyz3:M(1,1,1)§÷l=(2,−1,3)ê9FÝ.)µÏux=yz,uy=xz,uz=xy§K3(1,1,1):ux=uy=uz=1213qþl{ucosα=√,sinβ=−√,cosγ=√141414∂u2√K=ux(1,1,1)cosα+uy(1,1,1)cosβ+uz(1,1,1)cosγ=14¶gradu=i+j+k.∂l72223.¦êþ¼êu=x+2y+3z+xy+3x−2y−6z3O(0,0,0)9A(1,1,1)FÝ9Ù.)µÏux=2x+y+3,uy=4y+x−2,uz=6z−6K3O(0,0,0):µux=3,uy=−2,uz=−6§u´gradu=3i−2j+6k,|√gradu|=73A(1,1,1):µux=6,uy=3,uz=0§u´gradu=6i+3j,|gradu|=35.4.y²µ(1)grad(αu+βv)=αgradu+βgradv§Ù¥α,βÑ´~ê¶(2)grad(uv)=ugradv+vgrad课后答案网u¶0(3)gradF(u)=F(u)graduy²µ±¼ê~5y.-u=u(x,y),v=v(x,y)∂(αu+βv)∂u∂v∂(αu+βv)∂u∂v(1)Ï=α+β,=α+β∂x∂x∂x∂y!∂y∂y!!∂(αu+www.hackshp.cnβv)∂(αu+βv)∂u∂v∂u∂vKgrad(αu+βv)=,=α,+β,=αgradu+βgradv.∂x∂y∂x∂y∂x∂y∂(uv)∂u∂v∂(uv)∂u∂v(2)Ï=v+u,=v+u∂x∂x∂x∂y!∂y∂y!!∂(uv)∂(uv)∂u∂u∂v∂vKgrad(uv)=,=v,+u,=ugradv+vgradu.∂x∂y∂x∂y∂x∂y!!!∂F∂F∂u∂u∂u∂u0000(3)gradF(u)=,=F(u),F(u)=F(u),=F(u)gradu∂x∂y∂x∂y∂x∂yõ¼ê¼êy.1rp5.y²grad=−§r=x2+y2+z2,r=xi+yj+zk.rr3∂rx∂ry∂rzy²µÏ=,=,=∂xr∂y!r∂zr!1d11∂r∂r∂r11rKgrad=gradr=−i+j+k=−·(xi+yj+zk)=−.rdrrr2∂x∂y∂zr2rr3若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn2271p6.êþ¼êu=ln,r=(x−a)2+(y−b)2+(z−c)2§3m¥=:þ¤á|gradu|=1ºr∂rx−a∂ry−b∂rz−c)µÏ=,=,=∂xr∂y!r∂zr!d11x−ay−bz−c1Kgradu=lngradr=−i+j+k=−[(x−a)i+(y−b)j+(z−c)k]drrrrrrr21222u´|gradu|==1§Kr=1=(x−a)+(y−b)+(z−c)=1rùL²3±(a,b,c)¥%§»1¥¡þ¤á|gradu|=1.课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn228§7.Vúª221.Ñ:(1,−2)NC¼êf(x,y)=2x−xy−y−6x−3y+5Vúª.)µÏfx=4x−y−6,fy=−x−2y−3,fx2=4,fxy=−1,fy2=−2§¤kn þ0K3:(1,−2)§f=5,fx=0,fy=0,fx2=4,fxy=−1,fy2=−222u´f(x,y)=5+2(x−1)−(x−1)(y+2)−(y+2).x2.Ux9y¦Ðm¼êf(x,y)=eln(1+y)n.xxxeeexx)µÏfx=eln(1+y),fy=,fx2=eln(1+y),fy2=−2,fxy=1+y(1+y)1+yxxx2eeexfx3=eln(1+y),fy3=(1+y)3,fxy2=−(1+y)2,fyx2=1+yK3:(0,0)?§f=0,fx=fx2=fx3=0,fy=1,fxy=1,fy2=−1,fxy2=−1,fyx2=1,fy3=211112223u´f(x,y)=y+xy−y+xy−xy+y"2223课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn2291ÊÙ4Ú^4§1.4Ú¦{1.¦e¼ê4µ22(1)z=x−(y−1)2(2)z=(x−y+1)33(3)z=3axy−x−y(a>0)xπ(4)z=sinx+cosy+cos(x−y)066y2r22xy(5)z=xy·1−−(a,b>0)a2b2)µ(1)-zx=2x=0,zy=−2(y−1)=0Kx=0,y=1§u´:(0,1)U4:qzx2=2,zxy=0,zy2=−2§KA=2,B=0,C=−2§u´H=−4<0§ld¼êÃ4.(2)-zx=2(x−y+1)=0,zy=−2(x−y+1)=0K:©Ù3x−y+1=0þ§¼êUk4qA=2,B=−2,C=2§KH=0§I?ÚäÏéx−y+1=0þ:þkz=0§z>0ð¤áK¼êz3x−y+1=0þ:4z=0.22(3)-zx=3ay−3x=0,zy=3ax−3y=0x1=0x2=aK=3:(0,0),(a,a)?Uk4y1=0y2=aqzx2=−6x,zxy=3a,zy2=−6y2K3:(0,0)§A=0,B=3a,C=0§u´H=−9a<0§d¼êÃ4¶233:(a,a)§A=−6a<0,B=3a,C=−6a§u´H=27a>0§d¼êk4z=a.(4)-zx=cosx−sin(x−y)=0,zy=−siny+sin(x−y)=0!ππx3Kcosx=siny§u´y=−x§cosx−sin(x−y)=cos−sin2x−=2coscosx=0课后答案网2222!πx3ππππÏ06x6§Kcos6=0,cosx=0§u´x=,y=§=3:,?Uk42223636qzx2=−sinx−cos(x√−y),zxy=cos(x−y),zy2=−cosy−cos(x−!y)√3√9ππ3√KA=−3<0,B=,C=−3§u´H=>0§=3:,?¼êk4z=3.24362rwww.hackshp.cnr222222xyxyxyxy(5)-zx=y1−−−r=0,zy=x1−−−r=0a2b2x2y2a2b2x2y2a21−−b21−−a2b2a2b2aaaax2=√x3=−√x4=√x5=−√Kx1=03333y1=0bbbby2=√y3=−√y4=−√y5=√3!3!3!3!ababababu´3:P1(0,0),P2√,√,P3−√,−√,P4√,−√,P5−√,√?U4333333332223234424244422222244−3abxy+2bxy+3axyab−3abx+2bx−3aby+3abxy+2ayqzx2=3,zxy=3a4b21−x2−y22a4b41−x2−y22a2b2a2b22322233bxy−3abxy+2axyzy2=324x2y22ab1−a2−b2若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn232§2.^41.¦e¼ê3¤^e4µ22(1)f=x+y,ex+y=1¶222(2)f=x−2y+2z,ex+y+z=1¶1111(3)f=xyz,e++=(x>0,y>0,z>0,a>0)¶xyza11(4)f=+,ex+y=2¶xy222(5)f=xyz,ex+y+z=1,x+y+z=0.)µ22(1)¼êL=x+y+λ(x+y−1)√√22x1=x2=−Lx=1+2λx=0√2√222)§|Ly=1+2λy=0§y1=y2=−Lλ=x2+y2−1=02√√222λ1=−λ2=22qLx2=2√λ,Lxy√=0!,Ly2=2λ√√!22√22√222KdL,=−2(dx+dy)<0§u´¼ê3,?42¶2222√√!22√Ón§¼ê3−,−?4−2.22222(2)¼êL=x−2y+2z+λ(x+y+z−1)11x1=x2=−33Lx=1+2λx=022Ly1=−y2=y=−2+2λy=033)§|§Lz=2+2λz=022222Z1=z2=−Lλ=x+y+z−1=03333课后答案网λ1=−λ2=22qLx2=Ly2=Lz2=2λ,Lxy==Lxz=Lyz=0!1222222KdL(x2,y2,z2)=3(dx+dy+dz)>0§u´¼ê3−,,−?4−3¶333!122Ón§¼ê3,−,?43.333www.hackshp.cn!1111(3)¼êL=xyz+λ++−xyzaλLx=yz−=0x2λLy=xz−=0y24)§|§x=y=z=3a,λ=81aλLz=xy−=0z21111Lλ=++−=0xyzaqLx2(3a,3a,3a)=Ly2(3a,3a,3a)=Lz2(3a,3a,3a)=6a,Lxy(3a,3a,3a)=Lxz(3a,3a,3a)=Lyz(3a,3a,3a)=3a22222KdL(3a,3a,3a)=3a[(dx+dy+dz)+dx+dy+dz]>0§u´¼ê3(3a,3a,3a)?4327a.11(4)¼êL=++λ(x+y−2)xy若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn2331Lx=−+λ=0x2)§|1§x=y=λ=1Ly=−+λ=0y2Lλ=x+y−2=0qLx2(1,1)=Ly2(1,1)=2,Lxy(1,1)=0222KdL(1,1)=2(dx+dy)>0§u´¼ê3(1,1)?42.222(5)¼êL=xyz+u(x+y+z−1)+v(x+y+z)Lx=yz+2ux+v=0Ly=xz+2uy+v=0)§|Lz=xy+2uz+v=0§222Lu=x+y+z−1=0Lv=x+y+z=0√√√√√√666666x1=x2=−x3=−x4=x5=x6=−√6√6√33√6√√6666666y1=y2=−y3=y4=−y5=−y6=6√√6√6√6√33√666666z1=−√3z2=3√z3=√6z4=−√6z5=√6z6=−√6666666u1=u2=−u3=u4=−u5=u6=−121212121212111111v1=v2=v3=v4=v5=v6=6666662222qdL=2u(dx+dy+dz)+2(zdxdy+ydxdz+xdydz)√62222K3:(x1,y1,z1)?§dL=(dx+dy+dz−4dxdy+2dxdz+2dydz)6222dx+y+z=1§2xdx+2ydy+2zdz=0§K3:(x1,y1,z1)?§kdx+dy=2d√z22qdx+y+z=0§dx+dy+dz=0§Kdx=−dy,dz=0§u´dL(x1,y1,z1)=6dx>0§√√√!√6666K¼ê3,,−?4−66318√6Ón§¼ê3(x3,y3,z3),(x5,y5,z5)?4−√186¼ê3(x2,y2,z2),(x4,y4,z4),(x6,y6,z6)?4.18mnp2.¦f=xyz3^x+y+z=a,a>0,m>0,n>0,p>0,x>0,y>0,z>0e.mnp)µÏx>0,y>0,z>0§Kf=课后答案网xyz§lnf=mlnx+nlny+plnz§½,§I¦lnf4:§´f4:-L=mlnx+nlny+plnz+λ(x+y+z−a)mamx=Lx=x+λ=0m+n+pnnay=K)§Ly=y+λ=0§m+n+ppapwww.hackshp.cnz=Lz=+λ=0m+n+pzm+n+pLλ=x+y+z−a=0λ=−!amanapaK,,U4:m+n+pm+n+pm+n+p!mnpmnp2222qLx2=−x2,Lxy=Lyz=Lxz=0,Ly2=−y2,Lz2=−z2,dL=−x2dx−y2dy−z2dz<0!mnpmanapamnpm+n+p3,,?lnfk4§=fk4am+n+pm+n+pm+n+p(m+n+p)m+n+pmnpx+y=ax+z=ay+z=aqf=xnz(x,y,z)ªu>.§f→0§f4:z=0y=0x=0´§:.223.¦ýx+3y=12S n/§¦Ù.>²1uý¶§¡È.22xy)µduK¥n/Suý√+=1´ n/§.>²1u¶(23)24若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn234Ù.>¤éº:7´á¶þýº:(0,±2)n/,º:I(x,y)(x,y>0)§KÙS n/.>2x§py+2 n/nº:IA(0,−2),B(x,y),C(−x,y)§dýé¡5§A(0,2),B(x,−y),C(−x,−y)´Ùº:22KS=x(y+2)§:(x,y)3ýx+3y=12þ22qÏd¯K´¦S=x(y+2)3^x+3y=12þ(x,y>0)22¼êL=x(y+2)+λ(x+3y−12)x=3Lx=y+2+2λx=0y=1K)§Ly=x+6λy=0§221Lλ=x+3y−12=0λ=−2u´Ùº:IA(0,2),B(3,−1),C(−3,−1)½A(0,−2),B(3,1),C(−3,1)Ïd¯K¢S¯K§73§K3(0,2),(3,−1),(−3,−1)½(0,−2),(3,1),(−3,1)?١ȧÙ9.24.Á¦Ôy=4xþ:§¦§x−y+4=0åC.12)µ¤¦:I(x,y)§K§åld=√|x−y+4|§Ù¥y=4x2x−y+4=0ò²¡©!müÜ©§¡x−y+4<0§m¡x−y+4>012Ôy=4x3m¡Ü©§Ï:(x,y)§åld=√(x−y+4)212-L=√(x−y+4)+λ(y−4x)21Lx=√−4λ=0x=12y=2K)§|1§Ly=−√+2λy=012λ=√242Lλ=y−4x=021dy222qLx2=0,Ly2=√,Lxy=0,dL(1,2)=Lx2dx+2Lxydxdy+Ly2dy=√>02222(1,2)4:§=:(1,2)ålC.225.Ô¡z=x+y²¡xp+y+z=1 ¤ý§¦:ùýáål.)µâK¿§¦åld=x2+y2+z23^z=x2+y2,x+y+z=1222=¦u=x+y+z3^e2222课后答案网2L=x+y+z+λ(z−x−y)+γ(x+y+z−1)√√−1+3−1−3x1=x2=2√2√Lx=2x−2λx+γ=0−1+3−1−3Ly=2y−2λy+γ=0y1=y2=√2√2K)§|Lz=2z+λ+γ=0§z1=2−3,z2=2+322√√Lλ=z−x−y=0−53+953+9Lγ=x+y+z−1=0www.hackshp.cnλ1=λ2=3311√11√γ1=−7+3γ2=−7−3p√p√33u´d(x1,y1,z1)=9−53,d(x2,y2,z2)=9+53â¯K¢S¿Â§!áål3√√!1+31+3√p√Kål::−,−,2+3ål§9+53¶22√√!−1+3−1+3√p√áål::,,2−3ål§9−53.226.¦m:(a,b,c)²¡Ax+By+Cz+D=0áål.p)µ(x,y,z)²¡Ax+By+Cz+D=0þ?:§K§(a,b,c):åld=(x−a)2+(y−b)2+(z−c)2§Ù¥(x,y,z)÷vAx+By+Cz+D=02Ïd>0§d⇐⇒d2222UK§A¦d=(x−a)+(y−b)+(z−c)3^Ax+By+Cz+D=0e4222-L=(x−a)+(y−b)+(z−c)+λ(Ax+By+Cz+D)若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn2351x=a−λA2Lx=2(x−a)+λA=01Ly=b−λBK)§|y=2(y−b)+λB=0§2Lz=2(z−c)+λC=01z=c−λCLλ=Ax+By+Cz−D=022(Aa+Bb+Cc+D)λ=A2+B2+C2|Aa+Bb+Cc+D|u´d=√A2+B2+C2qx,y,z¥k?ªu∞§d→∞§Ïd3(x,y,z)(x−a)2+(y−b)2+(z−c)20,F(x0,y0−b)<0d^(1)§F(x,y)3«DþëY§Ï3η>0§¦|x−x0|<秽kF(x,y0+b)>0,F(x,y0−b)<0@"é∀x∈O(x0,η)§d¼êF(x,y)3[y0−b,y0+b]ëY59F(x,y0+b)>0,F(x,y0−b)<0â":3½n§73y∈(y0−b,y0+b)§¦F(x,y)=0duF(x,y)3[y0−b,y0+b]îüN§ly>y§F(x,y)>0¶y0(ε0,F(x1,y1−ε)<0KdFëY5§3O(x1,δ)⊂O(x0,a)§¦x∈O(x1,δ)§ðkF(x,y1+ε)>0,F(x,y1−ε)<0u´â":3½n§7ky∈O(y1,ε)§¦F(x,y)=0=y=f(x)=|x−x1|<δ§Òk|f(x)−f(x1)|=|y−y1|<ε=y=f(x)3x1:ëYdx1∈O(x0,a)?¿5§f(x)ëY¼ê.2222.¼êF(x,y)≡y−x(1−x)=03=:C/û½üëY§këYê¼êy=y(x).2223)µ¼êF(x,y)=y−x(1−x)3mëY§êFx=4x−2x,Fy=2yëY22222dy−x(1−x)=0§ey=0§Kx课后答案网(1−x)=0§)x=0,x=±1222qy>0,x>0§1−x>0=−16x61y6=0§Fy6=0dÛ¼ê3½n1§3?Û÷v{(x,y)|x|<1,x6=0,y2−x2(1−x2)=0}C/û½üëYkëYê¼êy=y(x).03.y²k¼êy=y(x)÷v§siny+sinhy=x§¿¦Ñêy(x).y²µ¼êF(x,y)=siny+sinhy−x3mëY§Fx=−1,Fy=cosy+coshyëYey+e−ywww.hackshp.cnqcoshy=>1 Ò3y=0¤á§dcosy=1§3¹e|cosy|612Ké:(x,y)§ðkFy=cosy+coshy>0§u´Fy6=0dÛ¼ê3½n1§3?Û÷vþã§:(x,y)§k¼ê÷v§siny+sinhy=xFx10Ù¼êy=−=.Fycosy+coshy4.D´:P0:(x0,y0,z0,u0,v0)§e(1)F(x0,y0,z0,u0,v0)=0,G(x0,y0,z0,u0,v0)=0¶(2)F,G"uCþê3D¥ëY¶FuFv(3)J=G3P0:Ø"¶uGvK3(x0,y0,z0)RS3é¼êu=f(x,y,z);v=g(x,y,z)÷vµ若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn237(1)u0=f(x0,y0,z0),v0=g(x0,y0,z0)(2)F(x,y,z,f,g)≡0,G(x,y,z,f,g)≡0(3)u=f(x,y,z),v=g(x,y,z)3:(x0,y0,z0)RSkéCþê§∂f1D(F,G)∂f1D(F,G)∂f1D(F,G)=−,=−,=−∂xJD(x,v)∂yJD(y,v)∂zJD(z,v)∂g1D(F,G)∂g1D(F,G)∂g1D(F,G)=−,=−,=−∂xJD(u,x)∂yJD(u,y)∂zJD(u,z)y²µd^(3)§Fu,Fv¥k3P0:Ø u0ØFv(P0)6=0§KdÛ¼ê3½n2§3P0:,S±rvlF(x,y,z,u,v)=0¥)Ñ5.v=ϕ(x,y,z,u)v0=ϕ(x0,y0,z0,u0)3(x0,y0,z0,u0),S´§äk"ux,y,z,uëYêrv=ϕ(x,y,z,u)G(x,y,z,u,v)¥G(x,y,z,u,ϕ(x,y,z,u))=ψ(x,y,z,u)!FuJψu=Gu+Gv·vu=Gu+Gv−=−FvFvdbFv(P0)6=03P0:J6=0§ψu(x0,y0,z0,u0)6=0Kd½n2§3(x0,y0,z0,u0),Sl§G=G(x,y,z,u,ϕ)≡ψ(x,y,z,u)=0¥)Ñu5.u=f(x,y,z)§3(x0,y0,z0),SkëYê§u0=f(x0,y0,z0)ru=f(x,y,z)ϕ(x,y,z,u)¥v=ϕ(x,y,z,f(x,y,z))=g(x,y,z)Kkg(x0,y0,z0)=ϕ(x0,y0,z0,u0)=v0u=f(x,y,z),v=g(x,y,z)=¤¦∂F∂F∂f∂F∂gF(x,y,z,u,v)=0∂x+∂u·∂x+∂v·∂x=0é§|üà"ux¦§G(x,y,z,u,v)=0∂G∂G∂f∂G∂g+·+·=0∂x∂u∂x∂v∂x∂f1D(F,G)∂g1D(F,G))§=−,=−∂xJD(x,v)∂xJD(x,u)∂f1D(F,G)∂f1D(F,G)∂g1D(F,G)∂g1D(F,G)Ón=−,=−,=−,=−∂yJD(y,v)∂zJD(z,v)∂yJD(u,y)∂zJD(u,z)5.ϕi(x)(i=1,2,···,n)´xëY¼ê§Gi(x1,···,xn)=Fi(ϕ1(x1),···,ϕn(xn))∂(G1,G2,···,Gn)Qn0y²=∆(ϕ(x1),···,ϕn(xn))ϕi(xi)∂(x1,x2,···,xn)i=1D(F1,F2,···,Fn)Ù¥∆(x1,x2,···,xn)=课后答案网D(x1,x2,···,xn)Qn0000ϕi(xi)=ϕ1(x1)ϕ2(x2)···ϕn(xn).i=1y²µÏϕi(x)(i=1,2,···,n)´xëY¼ê§Gi(x1,···,xn)=Fi(ϕ1(x1),···,ϕn(xn))∂Gi∂Fi∂ϕj∂Fi0K=·=ϕi(xj)(i,j=1,2,···,n)∂xj∂ϕj∂xj∂ϕj∂G1www.hackshp.cn∂G1∂G1∂F10∂F10∂F10···∂ϕϕ1(x1)ϕ2(x2)···ϕn(xn)∂x1∂x2∂xn1(x1)∂ϕ2(x2)∂ϕn(xn)∂G2∂G2∂G2∂F2∂F2∂F2∂(G1,G2,···,Gn)···ϕ10(x1)ϕ20(x2)···ϕn0(xn)u´=∂x1∂x2∂xn=∂ϕ1(x1)∂ϕ2(x2)∂ϕn(xn)=∂(x1,x2,···,xn).......................................................................................∂Gn∂Gn∂Gn∂Fn∂Fn∂Fn···000ϕ1(x1)ϕ2(x2)···ϕn(xn)∂x1∂x2∂xn∂ϕ1(x1)∂ϕ2(x2)∂ϕn(xn)∂F1∂F1∂F1···∂ϕ1(x1)∂ϕ2(x2)∂ϕn(xn)∂F2∂F2∂F2···QnD(F1,F2,···,Fn)ϕ0000=1(x1)ϕ2(x2)···ϕn(xn)∂ϕ1(x1)∂ϕ2(x2)∂ϕn(xn)=ϕi(xi)......................................i=1D(ϕ1(x1),ϕ2(x2),···,ϕn(xn))∂Fn∂Fn∂Fn···∂ϕ1(x1)∂ϕ2(x2)∂ϕn(xn)Qn0∆(ϕ(x1),···,ϕn(xn))ϕi(xi).i=16.F(x,y,z)k ëY꧿dF(x,y,z)=0(½z=f(x,y).?Øz=f(x,y)47Ú¿©^若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn238.2¦d222x+y+z−2x+2y−4z−10=0¤(½z=f(x,y)4.zx=fx(x,y)=0y²µÏ¼êz=f(x,y)47^zy=fy(x,y)=0FxFyFx=0qzx=−,zy=−§KF(x,y,z)47^FzFzFy=0qÛ¼ê4¿©^w¼êaÓ§´¦ ê^Û¼êp ê¦{Fx=2x−2=0x=1-)§éAzz1=−2,z2=6Fy=2y+2=0y=−12222222∂zx−1∂z1+y∂z(x−1)+(2−z)∂z(1+y)+(2−z)∂z(x−1)(1+y)q=,=,=,=,=∂x2−z∂y2−z∂x2(2−z)3∂y2(2−z)3∂x∂y(2−z)3222∂z1∂z1∂z1111u´3:(1,−1,−2),=,=,=0§d·−0=>09>0§Kz1=−24¶∂x24∂y24∂x∂y4!4!164222∂z1∂z1∂z11113:(1,−1,6),=−,=−,=0§d−−−0=>09−<0§Kz2=64∂x24∂y24∂x∂y44164课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn239§2.¼ê1ª5!¼ê"x=rcosθcosϕ1.y²y=rcosθsinϕ¼êÕáz=rsinθcosθcosϕ−rsinθcosϕ−rcosθsinϕD(x,y,z)y²µÏ=cosθsinϕ−rsinθsinϕrcosθcosϕ=−r2cosθD(r,θ,ϕ)sinθrcosθ0πD(x,y,z)K3r6=0θ6=kπ+«DS6=02D(r,θ,ϕ)u´â½n5§¼ê|3«DS¼êÕá.y1=x1+x2+···+xn2222.y²y2=x1+x2+···+xn¼ê"§¿ÑÙ¼ê"Xª.y3=x1x2+x1x3+···+xn−1xn1122y²µÏ3¼êϕ(t1,t2)=(t1−t2)§¦y3=ϕ(y1,y2)=(y1−y2)3nm(x1,x2,···,xn)S22¤ð ª12K¼ê|3nm¥¼ê"§Ù¼ê"Xªy3=(y1−y2).23.e¼ê´Ä"ºx−yy−zz−x(1),,x−zy−xz−yxyz(2),,1−x−y−z1−x−y−z1−x−y−z)µ(1)Ïf1·f2·f3=−1§K¼ê".xyz(2)-f1(x,y,z)=,f2(x,y,z)=,f3(x,y,z)=1−x−y−z1−x−y−z1−x−y−z∂f1∂f1∂f11−y−zxx∂x∂y∂z(1−x−y−z)2(1−x−y−z)2(1−x−y−z)2∂f2∂f2∂f2y1−x−zyKJocobiÝ∂x∂y∂z=(1−x−y−z)2(1−x−y−z)2(1−x−y−z)2∂f3∂f3∂f3zz1−x−y∂x∂y课后答案网∂z(1−x−y−z)2(1−x−y−z)2(1−x−y−z)2qdÝ3§Kâ½n6§¼ê|¼êÕá.www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn2401nÜ©¹ëCþÈ©Ú2ÂÈ©1ÔÙ¹ëCþÈ©Zy2−x2y01.F(y)=edx§OF(y).y)µÏ½n4^÷v§A^½n4§kZy22535533102−xy−y−y−y−yF(y)=(−x)edx+2ye−e=ye−e−F(y).y222yZy002.F(y)=(x+y)f(x)dx§Ù¥f(x)¼ê§¦F(y).0)µÏf(x)¼ê§KfZ(x)ëY§u´(x+y)f(x)ëY§K½n4^÷vy0000u´F(y)=f(x)dx+2yf(y)§F(y)=3f(y)+2yf(y).0Z1p3.eF(y)=lnx2+y2dx§OÈ©§¦ÑF(y)§2¦ÑF0(0)§¿uA^½n4OF0(0)(05.Z1Z21pp1xp)µy6=0§kF(y)=lnx2+y2dx=xlnx2+y2−dx=ln1+y2−1+x2+y2000xZ1dp1yy!dx=ln1+y2−1+yarctan.2y0x1+yZ1ÏF(0)=lnxdx=−10F(y)−F(0)πF(y)−F(0)π00KF+(0)=lim=,F−(0)=lim=−y→+0y2y→−0y20u´F(0)Ø3Z!∂py1∂p,¡§x>0§lnx2+y2==0§Klnx2+y2dx=0∂yx2+y20∂yy=0!y=0y=0!ZZπ1∂pπ1∂pqF+0(0)=6=0=lnx2+y2dx,F−0(0)=−6=0=lnx2+y2dx20∂y课后答案网20∂yy=0y=0Ky=0§ØU3È©Òe¦ê§=¦!mêØ1.4.¦¼êZπqZπ22dϕE(k)=1−k2sin2ϕdϕÚF(k)=p(00)x2+y20ëY5§Ù¥f(x)´[0,1]þëY¼ê.yf(x))µd>c>0§y∈[c,d]§Kȼê3[0,1]×[c,d]þëYx2+y2Z1yf(x)d½n1§F(y)=dx3[c,d]þëY§dc,d?¿5§F(y)3y>0ëYx2+y20qf(x)´[0,1]þëY¼ê§Kf(x)3[0,1]þ7km>0Z1y11πmπduF(y)>mdx=marctan9limarctan=§KlimF(y)>>00x2+y2yy→+0y2y→+02qF(0)=0§KF(y)y=0ØëY.6.A^éëê¦{OÈ©µZπ222(1)ln(a−sinx)dx(a>1)(Ø7½~ê§eOÑyÃ.¹§4O)¶0Zπ2(2)ln(1−2acosx+a)dx(|a|<1)0)µZπ222(1)I(a)=ln(a−sinx)dx0"#π22Ïȼêln(a−sinx)30,×[1,+∞]þØëY2"#πKØU^½n2§U^½n§0,×[b,c](b>1,c→+∞)2"#2aπ22ùf(x,a)=ln(a−sinx)9fa=Ñ34Ý/0,×[b,c]þëYa2−sin2x2Zπ"#2课后答案网2a2a+1a−1π0d½n2§kI(a)=dx=√arctan√+arctan√=√a2−sin2xa2−1a2−1a2−1a2−10√éaÈ©§I(a)=πln|a+a2−1|+C√Ïa∈[b,c]§db,c?¿5§KI(a)=πln|a+a2−1|+CZπ2(2)I(a)=ln(1−2acosx+a)dx0222|a|<1§du1−2acosxwww.hackshp.cn+a>1−2|a|+a=(1−|a|)>0−2cosx+2a2Kf(x,a)=ln(1−2acosx+a)9fa=Ñ34Ý/[0,π]×[−b,b]þëY(|a|6b<1)1−2acosx+a2Zπ−2cosx+2aπ1−a2Zπdxπ1−a2Zπdx0d½n2§kI(a)=dx=−=−01−2acosx+a2aa0(1+a2)−2acosxaa(1+a2)01+−2acosxa2+1xt=tanZ2Z22πdx+∞1+a1+aK=2dt=π01+−2acosx0(1−a)2+(1+a)2t21−a2a2+10u´I(a)=0§lI(a)=CqI(0)=0§KC=0§u´I(a)=0Ïa∈[−bb]§db?¿5§|a|<1§I(a)=0.7.A^È©Òe¦È©{OÈ©µZ!11xb−xasinlndx(a>0,b>0)0xlnx若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn2431lÙ¹ëCþ2ÂÈ©ZZ+∞+∞1.y²µe3[a,∞;c,d]S¤á|f(x,y)|6F(x,y)§¿"uy∈[c,d]È©F(x,y)dxÂñ§Kf(x,y)dx"aauy∈[c,d]½Âñ§ýéÂñ.Z+∞y²µÏÈ©F(x,y)dx"uy∈[c,d]Âñ§Kd¹ëCþ2ÂÈ©ÜÂñn§aZA00é∀ε>0§3yÃ"A0(ε)>a§A,A>A0§éy∈[c,d]§kF(x,y)dx<εAZA0ZA0ZA0f(x,y)dx6|f(x,y)|dx6F(x,y)dx<εéy∈[c,d]ѤáAAAZZ+∞+∞dá¹ëCþ2ÂÈ©ÜÂñn§f(x,y)dx"uy∈[c,d]Âñ§|f(x,y)|dx"aauy∈[c,d]ÂñZ+∞Kf(x,y)dx"uy∈[c,d]ÂñýéÂñ.a2.y²eÈ©3¤½«mSÂñµZ+∞cosxy(1)dx(y>a>0)x2+y20Z+∞cosxy(2)dx(−∞10by²µZcosxy1+∞dxπ(1)Ïy>a>0§K6=Âñx2+y2x2+a20x2+a22aZ+∞cosxyu´d¼O{§dx"uy3[a,+∞)(a>0)SÂñ.x2+y20Zcosxy1+∞dxπ(2)Ïy∈(−∞,+∞),6=Âñx2+1x2+10x2+12Z+∞课后答案网cosxyu´d¼O{§dx"uy3(−∞,+∞)SÂñ.x2+101b(3)x=0´Û:§6y6b,b>1,01)þÂñ.0bZ+∞3.f(x,y)3[a,+∞;c,d]ëY§é[c,d)þzy§f(x,y)dxÂñ§È©3y=duÑ.ay²ùÈ©3[c,d]Âñ.Z+∞ZA00000y²µdf(x,d)dxuѧ∃ε0>0,∀A0>a,∃A,A>A0§¦f(x,d)dx>ε0aA0ZA00Z+∞ùL²éy=d∈[c,d]kf(x,y)dx>ε0§`²f(x,y)dx3[c,d]Âñ.A0a4.?ØeÈ©3½«mÂñ5µZ+∞α−x(1)xedx(a6α6b;a,b?¿¢ê)1Z+∞√2−αx(2)αedx(0<α<+∞)0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn244Z+∞−(x−α)2(3)edx−∞(i)a<αp0>0(ii)p>0Z+∞−αx(5)esinxdx(α>0)0)µα−xb−x(1)Ïα∈[a,b],x∈(1,+∞)§K0<|xe|6xeZ+∞qlimx2·xbe−x=0§Kâá2ÂÈ©ÜO{4/ª§xbe−xdxÂñx→+∞Z1+∞α−xu´d¼O{§xedx"uα∈[a,b](a,b?¿¢ê)Âñ.1Z√+∞√2π−αx(2)αedx=Âñ§§3(0,+∞)"uαÂñ0Z2ZZ√+∞√2+∞2+∞2π−αx−t−té∀A>0§Ïlimαedx=limedt=edt=α→+0α→+0√2√AαA0ZZπ+∞√2+∞√2éu0<ε−α0x−α0x0<§73α0>0§¦α0edx=α0edx>ε0§Z2AA+∞√2−αx=αedx"uα3(0,+∞)þØÂñ.0ZZ+∞2+∞2√−(x−α)−(x−α)(3)é?¿½α∈(−∞,+∞)§È©edxÑÂñ§edx=π−∞−∞2x2−(x−α)−(i)|x|¿©§éa<α0§klim课后答案网edx=limedt=πα→+∞α→+∞A√A−αZ+∞2π−(x−α)Kα¿©§edx>ZA2+∞−(x−α)2d§edx3−∞<α<+∞þÂñZ0+∞−(x−α)2ledx3−∞www.hackshp.cn<α<+∞þÂñ.−∞(4)(i)|xp−1ln2x|=xp−1ln2x6xp0−1ln2x(p>p0>0,06x61)ZZ1+∞È©xp−1ln2xdx=e−p0zz2dz004zlimz2·e−p0zz2=lim=0(p0>0)z→+∞z→+∞Zep0zZ+∞1KdÜO{4/ªe−p0zz2dzÂñ§u´xp0−1ln2xdxÂñZ001p−12ld¼O{§xlnxdx"up3p>p0>0þÂñ.!012(ii)Ïx∈0,,lnx>1eZZ1Z1!p1p−12ep−12ep−111kxlnxdx>xlnxdx>xdx=→+∞(p→+0)000peZ1p−12u´xlnxdx3p>0ØÂñ.0若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn258)¶(1)|^ΡIx=rcosθ,y=rsinθ,z=z|J|=r§3dCe§¡§Cµ222r+z=4R,r=2rcosθ√!ZZZZπZ2RcosθZ4R2−r221634KV=dxdydz=2dθrdrdz=Rπ−−π0033V2(2)dKáN31!1n!1891l%S§éuù%©Okx,y,z>0;x,y60,z>0;x,z60,y>0;x>0,y,z60Ϫà9màx,y,z¥?üÓCÒ ªE¤á§áN3ùo%SÜ©§éé/é¡uI¶.2d¥¡Ix=ρsinϕcosθ,y=ρsinϕsinθ,z=ρcosϕ,|J|=ρsinϕ63232¡§Cµρ=3ρsinϕcosϕsinθcosθ=ρ=3sinϕcosϕsinθcosθ§ππ31%S§ρ>0,06θ6,06ϕ62√2ZπZπZ33sin2ϕcosϕsinθcosθ2221u´V=4dθsinϕdϕρdρ=.00025.|^·ICOe¡¤Nȵ!222222xyzxy(1)++=+a2b2c2a2b2!2!2xyz(2)++=1,(x>0,y>0,z>0,a,b,c>0)abc222222(3)z=x+y,z=2(x+y),xy=a,xy=2a,x=2y,2x=y§(Ù¥x,y>0))¶(1)d2Â¥¡I:x=aρsinϕcosθ,y=bρsinϕsinθ,z=cρcosϕ§Ù¥ρ>0,06θ62π,06ϕ6π§2ù|J|=abcρsinϕ¡§Cµρ=sinϕZ2πZπZsinϕπ22KV=abcdθsinϕdϕρdρ=abc0004ππ22(2)Cµx=arcosθcosϕ,y=brsinθcosϕ,z=crsinϕ§Ù¥06r61,06θ6,06ϕ6§222ù|J|=2abcrcosθsinθcosϕZπZπZ1222abcKV=dθdϕ(2课后答案网abcrcosθsinθcosϕ)dr=0003r!√vv22(3)-z=u(x+y),xy=v,x=wy§Kx=wv,y=,z=uwv+wwvv122d|J|=+§16u62,a6v62a,6w6222w2!2Z2Z2a2Z2119www.hackshp.cn4u´V=duvdv+dw=a21a2122w426.¦äküNÈ06x61,06y61,06z61ÔNþ§eÔN3:M(x,y,z)ݵ=x+y+z.ZZZZZZ1113)¶m=(x+y+z)dxdydz=dxdy(x+y+z)dz=.0002V若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn259§3.È©3ÔnþA^1.¦e¤. %Iµ2(1)ay=x,x+y=2a(a>0)(2)r=a(1+cosϕ)(06ϕ6π))µRRRRxdΩydΩΩΩ(1)Ýρ~ê§Kx=RR,y=RRGdΩGdΩΩΩZZZZa2a−x92ddΩ=dxdy=a−2ax22ΩaZZZZa2a−x93xdΩ=xdxdy=−a−2ax24ΩaZZZZa2a−x363ydΩ=dxydy=a−2ax25Ωaa8Kx=−,y=aG2G5RRRRrcosϕdΩrsinϕdΩΩΩ(2)Ýρ~ê§Kx=RR,y=RRGdΩGdΩΩΩZZZZπa(1+cosϕ)32ddΩ=dϕrdr=aπ004ΩZZZZπa(1+cosϕ)523rcosϕdΩ=dϕrcosϕdr=aπ008ΩZZZZπa(1+cosϕ)423rsinϕdΩ=dϕrsinϕdr=a003Ω516aKx=a,y=.G6G9π2.¦de¡¤.ÔN%µ222xyz课后答案网(1)++=1,x>0,y>0,z>0a2b2c222(2)z=x+y,x+y=a,x=0,y=0,z=0)µRRRRRRRRRxdxdydzydxdydzDzdxdydzVVV(1)Ýρ~ê§Kx=RRR,y=RRR,z=RRRGdwww.hackshp.cnxdydzGdxdydzGdxdydzVVVππ-x=aρsinϕcosθ,y=bρsinϕsinθ,z=cρcosϕ§Ù¥06ρ61,06θ6,06ϕ6§d222|J|=abcρsinϕZZZZπZπZ1222πKdxdydz=abcdθsinϕdϕρdρ=abc0006VZZZZπZπZ122223π2xdxdydz=abccosθdθsinϕdϕρdρ=abc00016V333u´x=a§dé¡5§y=b,z=cG8G8G8RRRRRRRRRxdxdydzydxdydzzdxdydzVVV(2)Ýρ~ê§Kx=RRR,y=RRR,z=RRRGdxdydzGdxdydzGdxdydzVVVZZZZaZa−xZx2+y24addxdydz=dxdydz=0006V若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn260ZZZZaZa−xZx2+y25axdxdydz=xdxdydz=00015VZZZ5ZZZa76ydxdydz=,zdxdydz=a15180VV2272Kx=a,y=a,z=a.G5G5G303.¦þ!©Ùuür=2sinθ9r=4sinθm«þþ%.)µdé¡5§x=0!ZZ1π4sinθ772qy=dθrsinθdr=§K¤¦/%0,.3π02sinθ334.3,)L§¥§3/>þVþ>» Ý/§¦²¡ã/%á3 %þ§Á¦Ý/,>.)µÝρ~ê§Ý/,>l§%3I:(0,0)§ux¶þ§Ý/ux¶e√ZRZR2−x2ρxdxdy−R−lu´x= =0ρ1πR2+2Rl2√ZRZR2−x2ρdxydy!22−R −l==22y=R−lρ1πR2+2RlπR+4l32√6-y=0§Kl=R.325.¦þ!©Ù3dy=xy=1¤¤²¡ã/þþ"uy=−1=Ä.þ.ZZZZ1136822)µIy=1=(y+1)dΩ=dx(y+1)dy=.−1x2105Ω6.¦de¡¤.þ!Néu¤«¶=Ä.þµ22(1)z=x+y,x+y=±1,x−y=±1,z=0"uz¶(2)N"u§c.)µ(1)¡¤.þ!ÔNéuOZZZZ¶=Ä.þPIOZKI22课后答案网OZ=(x+y)dxdydzVZ1Z1−xZx2+y2Z0Zx+1Zx2+y2142222=dxdy(x+y)dz+dxdy+(x+y)dz=0x−10−1−(1+x)045(2)N06z6c,06y6b,06x6aZZZabcabc2222"uz¶=Ä.þIOZ=www.hackshp.cndxdy(x+y)dz=(a+b).00032227.¦þ! ¡x+y6R,z=0éuz¶þ:(0,0,c)(c>0)?üþÚå.)µÚå3OX,OY¶þK0§=Fx=Fy=0§ρ=ρ0"#ZZZZc2πRcr11KFz=kρ0dΩ=kρ0dθdr=2kρ0πc−√.d3223cR2+c200(r+c)2Ω2228.¦þ!ÎNx+y6a,06z6héup(0,0,c)(c>h):?üþÚå.)µρ=ρ0§dé¡5§Úå3OX,OY¶þK0§=Fx=Fy=0|^ΡI§Úå3OZ¶þKµZZZZZZhz−c2πahz−cFz=kρ0dxdydz=kρ0dθrdrdz330(x2+y2+(z−c)2)2000[r2+(z−c)2]2pΩp=2πkρ0(a2+c2−a2+(c−h)2−h).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn261§4.2ÂÈ©1.Oe2ÂÈ©µZZdxdy(1)xpyqxy>1x>1ZZdxdy(2)p1−x2−y2x2+y261ZZ+∞+∞−(x2+y2)(3)edxdy.¿dy²VÇÈ©−∞−∞Z1+∞2−x√edx=1π−∞)µZZZZdxdy+∞dx+∞dy(1)duȼêK§I==pqpqxy1x1yxy>1xx>1ZZ1+∞dy+∞dyq61§dx>1§0<61§KÈ©uÑk=+∞§qqx1y1yZZxxdxdyu´I==+∞xpyqxy>1x>1Z+∞dyxq−1q>1§=q1yq−1xZZZ+∞dx+∞dy1+∞1q−p−1d§p>q§I==xdx=pq1x1yq−11(q−1)(p−q)xZZZ1+∞dxdyq−p−1p6q§(p+1)−q61§KÈ©xdx=+∞§lI==+∞q−11xpyqxy>1x>1ZZdxdy1nþ§p>q>1§I==§Ù{¹I=+∞.课后答案网xpyq(q−1)(p−q)xy>1x>1ZZZZdxdy2πεr(2)p=limdθ√dr=2π.1−x2−y2ε→1001−r2x2+y261(3)Cx=rcosθ,y=rsinθ(06θ62π,r>0),|J|=rZZZZ+∞+∞www.hackshp.cn2π+∞−(x2+y2)−r2Kedxdy=dθredr=π.Z−∞Z−∞Z00ZZZ+∞+∞+∞+∞+∞+∞−(x2+y2)−x2−y2−x2−y2dedxdy=edxedyedx=edy,−∞−∞−∞−∞−∞−∞Z+∞2√1R+∞2−x√−xKedx=π=edx=1.π−∞−∞2.?Øe¡2ÂÈ©Âñ5µZZ+∞+∞dxdy(1)−∞−∞(1+|x|p)(1+|y|q)ZZϕ(x,y)(2)dxdy,01§È©Âñ¶p61§È©uÑ01+xp01+xpZZ+∞dy+∞dyÓn§q>1§È©Âñ¶q61§È©uÑ01+yq01+yqZZ+∞+∞dxdynþ§p>1q>1§È©Âñ§Ù{¹þuÑ.−∞−∞(1+|x|p)(1+|y|q)mϕ(x,y)M(2)Ï66(1+x2+y2)p(1+x2+y2)p(1+x2+y2)pZZZZϕ(x,y)dxdyKd2ÂÈ©"O{92ÂÈ©5§dxdy(1+x2+y2)p(1+x2+y2)p06y6106y61ÓñÑZZZZdxdy1+∞dxdȼêé¡59K5§=2dy(1+x2+y2)p00(1+x2+y2)p06y61du06y61§KZZZ+∞dx+∞dx+∞dxep>0§K660(2+x2)p0(1+x2+y2)p0(1+x2)pZZZ+∞dx+∞dx+∞dxep<0§K>>0(2+x2)p0(1+x2+y2)p0(1+x2)PZ1+∞dx112péuα>0§dulimx=1§KÈ©p>Âñ¶p6uÑx→+∞(α2+x2)p0(α2+x2)p22ZZdxdy11u´p>Âñ¶p6uÑ(1+x2+y2)p2206y61ZZϕ(x,y)11ldxdyp>Âñ¶p6uÑ.(1+x2+y2)p课后答案网2206y61mϕ(x,y)M(3)Ï0<66|x−y|p|x−y|p|x−y|pZZZZaaϕ(x,y)aadxdyKd2ÂÈ©"O{92ÂÈ©5§dxdyÓñÑ00|x−y|p00|x−y|pZZZZwww.hackshp.cnaadxdyaxdydȼêé¡59K5§dxdy=2dx00|x−y|p00(x−y)pZaZxdya2−pp<1§dx=00(x−y)p(2−p)(1−p)ZaZadxdyZaZxdy2a2−pK=2dx=00|x−y|p00(x−y)p(2−p)(1−p)ZZZZaxdyax−εdyp>1§dx=limdx00(x−y)pε→+0ε0(x−y)pZZZZaxdyax−εdyp=1§dx=limdx=lim(alna−a+ε−alnε)=+∞00(x−y)pε→+0ε0x−yε→+0ZZaadxdyKp=1uѶ00|x−y|p!ZZZZaxdyax−εdya+εlnεp=2§dx=limdx=lim−1−lna=−∞00(x−y)pε→+0ε0(x−y)2ε→+0εZZaadxdyKp=2uѶ00|x−y|p若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn263ZZZZaxdyax−εdyp>1p6=2§dx=limdx"00(x−!y)pε→+0ε0#(x−y)p1p−11=lima−ε+=+∞ε→+0(p−1)εp−1p−2(p−1)(p−2)ap−2ZZaadxdy)Kp>1p6=2uÑ00|x−y|pZZaaϕ(x,y)nþdxdyp<1Âñ¶p>1uÑ.00|x−y|p22(4)(0,0)´Û:§dux++xy+y>0((x,y)6=(0,0))§Kmϕ(x,y)M66(x2+xy+y2)p(x2+xy+y2)p(x2+xy+y2)pZZZZaaϕ(x,y)aadxdyd2ÂÈ©"O{92ÂÈ©5§dxdy00(x2+xy+y2)p00(x2+xy+y2)pÓñÑZZZZZdxdy1dr2πdθ1dr=lim=Nlim(x2+xy+y2)pε→+0εr2p−10(1+sinθcosθ)pε→+0εr2p−1x2+y261Nlim(−lnε)=+∞,p=1!ε→+0Z2π=2−2pNÙ¥N=dθ~ÂÈ©§~þ1−ε,p<1(1+sinθcosθ)pNε→lim+02−2p=2−2p0∞,p>1ZZdxdyo§p<1Âñ¶p>1uÑ(x2+xy+y2)px2+y261ZZϕ(x,y)ldxdyp<1Âñ¶p>1uÑ.(x2+xy+y2)px2+y261ZZdxdy2223.y²D´d31Ôy=x§±x+y=19x¶¤¤«§K3.x2+y2Dy²µ(0,0)´Û:ZZdxdyZθ0Z1drZθ0cos2θ=dθ=lndθ,0´Û:22x+y0sinθr0sinθDcos2θ!√sinθ05−1Ù¥θ0÷v2=1=sinθ0=课后答案网cosθ02cos2θZθ0cos2θ1Ïlimθ2ln=0§KdÜO{4/ª§lndθÂñθ→+0ZZsinθ0sinθdxdylÈ©3.x2+y2Dwww.hackshp.cn4.¦þ!IN"u3§º:?þm:Úå.)µÚå3OX,OY¶þK0§=Fx=Fy=0§ZZZZZZh2πRρmzmRρz2mRπFz=dV=dθdρdz=(l−g).33grg000(ρ2+z2)2glV若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn268§3.1aÈ©1.Oe1aÈ©µZ222(1)(x−2xy)dx+(y−2xy)dy,ly=xl(1,1)(−1,1)¶lI2222(2)(x+y)dx+(x−y)dy,l±A(1,0),B(2,0)C(2,1),D(1,1)º:/§¶lZ(3)(2a−y)dx+dy,l^Óx=a(t−sint),y=a(1−cost)§l(0,0)(2π,0)¶lZ(4)ydx−xdy+(x2+y2)dz,lx=et,y=e−t,z=atl(1,1,0)(e,e−1,a)l)µZZ−114222343(1)(x−2xy)dx+(y−2xy)dy=[x−2x+2x(x−2x)]dx=.l115IZZZZ22222222(2)I=(x+y)dx+(x−y)dy=+++(x+y)dx+(x−y)dylZABBCCDDA22÷AB§y=0§(x−y)dy=0ZABZZ222222Ó§k(x+y)dx=(x−y)dy=(x+y)dx=0ZBCZCDZDAZ21102222KI=xdx+(4−y)dy+(x+1)dx=(1−y)dy=2.1021ZZ2π2(3)(2a−y)dx+dy=[(a+acost)·a(1−cost)+asint]dt=aπ.l0ZZ1a22−ttt−t2t−2t2−2(4)ydx−xdy+(x+y)dz=[e·e−e(−e)+(e+e)a]dt=2+(e−e).l022.¦È©Z(1,1,1)ijkJ=111·dr(0,0,0)xyzÙ¥dr¥»§È©´»©Oµ(1)÷¶!2ϕπ(2)÷r=isinϕ+j(1−cos课后答案网ϕ)+k,06ϕ6.π2)µ(1)§µx=y=zZ(1,1,1)ijkZ(1,1,1)KJ=111·dr=(z−y)dx+(x−z)dy+(y−x)dz=(0,0,0)xyz(0,0,0)Z1Z1www.hackshp.cnZ1(x−x)dx+(y−y)dy+(z−z)dz=0.000ZZ("#!)(1,1,1)ijkπ22ϕ2ϕ2(2)J=111·dr=−(1−cosϕ)cosϕ+sinϕ−sinϕ+[(1−cosϕ)−sinϕ]·dϕ=(0,0,0)xyz0ππππ81−−.2π3.1w4L31w¡Sþ§S§z=f(x,y)§L3XY¡þÝKl§¼êP(x,y,z)3LþëY§y²IIP(x,y,z)dx=P[x,y,f(x,y)]dxLly²µØS¡þý§z=f(x,y),(x,y)∈DK¡>.L3XY²¡þÝKA´_l:x=ϕ(t),y=ψ(t),aF/abPSÚabQP¡È(Ù¥a,b©SP,RQX¶:)Zbu´kS=[y1(x)−y0(x)]dxaZZZZbb,¡§âII.Oúª§kydx=y0(x)dx,ydx=y1(x)dx__ZZwww.hackshp.cnPQaSRa¿5¿ydx=ydx=0PSPQZZZZZZZZbbK−ydx=ydx=+++ydx=y1(x)dx−y0(x)dx=S__LZPSRQPPSSRRQQRaa=S=−ydx.LZ(2)éu«D=PQRS§Ù¥PQ,RSkX¶§Ón§kxdy=S.L(3)éuE,«/zþü«/§ÓOì¡È§, §5¿E´p-§Óþü«(J.I1nþ¤ã§kS=xdy−ydx.2L6.Oe¤«¡Èµ(1)ýµx=acost,y=bsint,(06t62π)¶33(2)(/µx=acost,y=asint,(06t62π).若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn27422(2)y=tx§K§zx(1+t)=ax(a>0,x>0)aatu´ëê§x=,y=(06t<+∞)(1+t)2(1+t)2§Ox¶:(a,0)(0,0)2a3Ox¶þl(0,0):(a,0):ãþkxdy−ydx=0¶3Ôþkxdy−ydx=dt(1+t)41Ia2Z+∞dta2u´¡ÈD=xdy−ydx==.2l20(1+t)463.y²eC²¡þµ4§l?¿KIcos(l,n)ds=0cª¥nC{.y²µØC_Ï(l,n)=(l,x)−(n,x)§Kcos(l,n)=cos(l,x)cos(n,x)+sin(l,x)sin(n,x)dxdyqsin(n,x)=−cos(t,x),cos(n,x)=sin(t,x)cos(t,x)=,sin(t,x)=dsdsKcos(lI,n)ds=cos(l,xI)dy−sin(l,x)dxu´cos(l,n)ds=[−sin(l,x)dx+cos(l,x)dy]CC∂P∂QdP=−sin(l,x),Q=cos(l,x)§=0=∂y!∂xIZZ∂Q∂Pu´cos(l,n)ds=−dxdy=0C∂x∂yD4.u(x,y),v(x,y)´äk ëYê¼ê§¿22∂u∂u∆u≡+∂x2∂y2y²µZZZ∂u(1)∆udxdy=dsl∂nσZZZZ!I∂u∂v∂u∂v∂u(2)v∆udxdy=−+dxdy+vds∂x课后答案网∂x∂y∂yl∂nσσZZZ!∂u∂v(3)(u∆v−v∆u)dxdy=−v−udsl∂n∂nσ∂u∂vÙ¥σ4l¤²¡«§,÷l{ê.∂n∂ny²µwww.hackshp.cnZZ!ZZ∂u∂u∂u∂u∂u(1)ds=cos(n,x)+sin(n,x)ds=sin(t,x)ds−cos(t,x)dsl∂nl∂x∂yl∂xl∂yZZZZ"22#ZZ∂u∂u∂u∂u=dy−dx=+dxdy=∆udxdyl∂xl∂y∂x2∂y2σσII!I"#∂u∂u∂u∂u∂u(2)Ïvds=vcos(n,x)+sin(n,x)ds=vsin(t,x)−vcos(t,x)dsl∂nl∂x∂yl∂x∂yIZZ"!!#∂u∂u∂∂u∂∂u=vdy−vdx=v−−vdxdyl∂x∂y∂x∂x∂y∂yZZ!σZZ∂u∂v∂u∂v=+dxdy+v∆udxdy∂x∂x∂y∂yZZσZZσ!I∂u∂v∂u∂v∂uKv∆udxdy=−+dxdy+vds∂x∂x∂y∂yl∂nσσ若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn276(1)ÏP=yz,Q=xz,RZZ=xy§KPx=Qy=Rz=0u´dpdúª§xydxdy+xzdxdz+yzdydz=0S(2)ÏP=ux,Q=uy,R=uz§Kdpdúª§ZZ!ZZZ222!∂u∂u∂u∂u∂u∂ucosα+cosβ+cosγdS=++dxdydz∂x∂y∂z∂x2∂y2∂z2SV8.|^pdúªO¡È©µZZ222(1)xdydz+ydxdz+zdxdy,SµáN06x,y,z6aL¡¶SZZ333(2)xdydz+ydxdz+zdxdy,Sµü¥L¡¶SZZ222222(3)(xcosα+ycosβ+zcosγ)dS,Sµx+y=z,06z6h;cosα,cosβ,cosγd¡{S{u.)µ222(1)ÏP=x,Q=y,RZZ=z§ZZZaaa2224Kdpdúª§xdydz+ydxdz+zdxdy=2dxdy(x+y+z)dz=3a000S333(2)ÏP=x,Q=y,RZZ=zZZZ333222Kdpdúª§xdydz+ydxdz+zdxdy=3(x+y+z)dxdydzSVZZZ2ππ1124=3dθsinϕdϕrdr=π0005(3)dpdúª§ZZZZZ222(xcosα+ycosβ+zcosγ)dS=2(x+y+z)dxdydzSVZ2πZhZhπh4=2dϕrdr[r(cosϕ+sinϕ)+z]dz=.00r29.y²µe课后答案网∂2u∂2u∂2u∆u≡++∂x2∂y2∂z2S´V>.¡§K¤áe¡úªµZZZZZ∂u(1)∆udxdydz=dS∂nVSZZZZZ"!www.hackshp.cn2!2!2#ZZZ∂u∂u∂u∂u(2)udS=++dxdydz+u∆udxdydz∂n∂x∂y∂zSVV∂uª¥u3V+SþkëY ê§÷¡S{ê.∂ny²µZZZZ"#ZZZ"222#∂u∂u∂u∂u∂u∂u∂u(1)dS=cos(n,x)+cos(n,y)+cos(n,z)dS=++dxdydz=∂n∂x∂y∂z∂x2∂y2∂z2SZZZSV∆udxdydzVZZZZ"#∂u∂u∂u∂u(2)udS=ucos(n,x)+ucos(n,y)+ucos(n,z)dS∂n∂x∂y∂zSZZZ"S!!!#∂∂u∂∂u∂∂u=u+u+udxdydz∂x∂x∂y∂y∂z∂zV若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
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课后答案网：www.hackshp.cn2792222(1)(x+2xy−y)dx+(x−2xy−y)dy22(2)(2xcosy−ysinx)dx+(2ycosx−xsiny)dyab−by−ax(3)dx+dy+dzzzz2222(4)(x−2yz)dx+(y−2xz)dy+(z−2xy)dzxyxy(5)e[e(x−y+2)+y]dx+e[e(x−y)+1]dy)µ∂P∂Q∂P∂Q(1)Ï=2x−2y,=2x−2y§K=∂y∂x∂y∂xZxZyx3y322222u´u=(x+2xy−y)dx+(−y)dy+C=+xy−xy−+C0033222222(2)Ï(2xcosy−ysinx)dx+(2ycosx−xsiny)dy=cosydx+xdcosy+ydcosx+cosxdy=22d(xcosy+ycosx)22Ku=xcosy+ycosx+Cab−by−axzdx−xdzzdy−ydzax+by(3)Ïdx+dy+dz=a+b=dzzz2z2z2z1Ku=(ax+by)+Cz1222333(4)Ï(x−2yz)dx+(y−2xz)dy+(z−2xy)dz=(dx+dy+dz)−2(yzdx+xzdy+xydz)=!3333x+y+zd−2xyz31333Ku=(x+y+z)−2xyz+C3xyxyx+yx+yx(5)Ïe[e(x−y+2)+y]dx+e[e(x−y)+1]dy=(x−y)e(dx+dy)+2edx+d(ye)=x+yx+yxx+yxd((x−y)e)+ed(x+y)+d(ye)=d((x−y+1)e)+d(ye)x+yxKu=(x−y+1)e+ye+C4.yµ1xdy−ydxPdx+Qdy=2Ax2+2Bxy+Cy22(A,B,C~ê§AC−B>0)·Ü^µ∂P∂Q课后答案网=∂y∂x¦µ"uÛ:(0,0)Ì~ê.yxy²µÏP=−,Q=2(Ax2+2Bxy+Cy2)2(Ax2+2Bxy+Cy2)22∂PCy−Ax∂QK==∂y2(Ax2+2Bxy+Cy2)2∂xIZwww.hackshp.cnπ Bπ22dt1Ctant+Cu´ω=Pdx+Qdy==√arctan√2+y2=1π2(Acos2t+2Bsintcost+Csin2t)AC−B2AC−B2x−π2−2π√,C>0=AC−B2π−√,C<0AC−B25.y²µZxdx+ydyx2+y2xdx+ydy"uÛ:(0,0)Ì~ê0§lÈ©´»Ã".x2+y2xy∂P2xy∂Qy²µÏP=,Q=§K=−=Ix2+y2x2+y2∂y(x2+y2)2∂xu´ω=Pdx+Qdy=0x2+y2=1若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn280(1)e4´lØ(0,0):§òÛ:(0,0)«D>.^^Cëå5§u´EëÏ«C¤üëÏ«I∂P∂Qxdx+ydyq=§Kd d^§=0∂y∂xlx2+y2Ixdx+ydy(2)e4´lÛ:(0,0)§Ï÷7Û:?4´È© uÌ~ê§K=0x2+y2lxdx+ydyoÈ©´»Ã".x2+y2课后答案网www.hackshp.cn若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn281§3.|ØÐÚt−t1.H(t)=ea+eb§Ù¥a,b~þ§tëê§dH(1)¦dt2dH(2)y²=Hdt2t−t)µÏH(t)=ea+eb§a,b~þ§KdHt−t(1)=ea−ebdt!2dHddHt−t(2)==ea+eb=Hdt2dtdt!!ddAdBdC2.y²µ[A·(B×C)]=·(B×C)+A·×C+A·B×dtdtdtdty²µA=Axi+Ayj+Azk,B=Bxi+Byj+Bzk,C=Cxi+Cyj+CzkAxAyAzKA·(B×C)=BxByBz§é ªüধmà^é1ª¦{K§CxCyCzAxAyAzAxAyAzAxAyAzdttt[A·(B×C)]=BxByBz+BxtBytBzt+BxByBzdtCxCyCzCxCyCzCxCyCz!!tttdAdBdC=·(B×C)+A·×C+A·B×dtdtdt3.a=3i+20j−15k§éeêþ|φ©O¦Ñgradφ9div(φa).222−1(1)φ=(x+y+z)2222(2)φ=x+y+z222(3)φ=ln(x+y+z))µ−x−y−z(1)gradφ=φxi+φyj+φzk=i+j+k333课后答案网(x2+xy+y2)2(x2+xy+y2)2(x2+xy+y2)2−3x−20y+15zdiv(φa)=φdiva+gradφ·a=gradφ·a=3(x2+xy+y2)2(2)gradφ=2xi+2yj+2zk,div(φa)=6x+40y−30z2x2y2z6x+40y−30z(3)gradφ=i+j+k,div(φa)=x2+y2+z2x2+y2+z2x2+y2+z2x2+y2+z24.U(x,y,z)=xyzwww.hackshp.cn(1)¦U(x,y,z)3:P1(0,0,0),P2(1,1,1)9P3(2,1,1)?÷b=2i+3j−4kê¶(2)3þãn:?§U(M)êÛº(3)3þãn:?§¦divgradU(M)9rotgradU(M).)µ234(1)Ïb{ucosα=√,cosβ=√,cosγ=−√292929∂U1K=yzcosα+xzcosβ+xycosγ=√(2yz+3xz−4xy)∂b29√∂U∂U29∂Uu´3P1(0,0,0):=0¶3P2(1,1,1):=¶3P3(2,1,1):=0∂b∂b29∂b∂U(2)Ï=gradU·b0=|gradU|cos(gradU,b0)§Ù¥b0´büþ∂bpKU(M)ê|gradU|=y2z2+x2z2+x2y2√u´3P1(0,0,0):|gradU|=0¶3P2(1,1,1):|gradU|=3¶3P3(2,1,1):|gradU|=3若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn282(3)ÏgradU=yzi+xzj+xyk∂(yz)∂(xz)∂(xy)KdivgradU=++=0∂x∂y!∂z!!∂(xy)∂(xz)∂(yz)∂(xy)∂(xz)∂(yz)rotgradU=−i+−j+−k=0∂y∂z∂z∂x∂x∂yu´3þãn:?§divgradU(M)=0,rotgradU(M)=0.2222225.¦þa=xi+yj+zkBL¥x+y+z=1,x>0,y>0,z>06þ.ZZZZZπZ1222222π)µΦ=xdydz+ydxdz+zdxdy§zdxdy=dθ(1−r)rdr=008SSZZZZπ322aq/§©OXOZ,YOZ²¡ÝK§ydxdz=xdydz=§u´Φ=π.88SS6.¦a=yzi+xzj+xykÏLS6þ§222(1)SÎNx+y6a,06z6hý¡¶(2)S(1)¥ÎNþ.¡¶(3)S(1)¥ÎNL¡.)µZZZZZZZZ"#∂(yz)∂(xz)∂(xy)(3)andS=divadV=++dV=0∂x∂y∂zSVVu´þaBLÎNL¡6þ0ZZZZZZ2πa3(2)Ï3 ÎNþ!e.¡an=xy§KandS=xydxdy=dθrsinθcosθdr=000Sþx2+y26a2ZZÓnandS=0§u´þaBLÎNþ.¡6þ0ZZSeZZZZZZZZ(1)ÏandS=andS+andS+andS§KandS=0.SSýSþSeSý!y7.¦a=gradarctan÷l6þµx22(1)l(x−2)+(y−2)=1,z课后答案网=0¶22(2)lx+y=4,z=1.)µ!yyx(1)d®§ka=gradarctan=−i+jxx2+y2x2+y2xy∂x2+y2∂−www.hackshp.cnx2+y2Krota=−k=0(Øx=y=0=Oz¶þ:)∂x∂y22Ïl:(x−2)+(y−2)=1,zI=0´Ø7zZZ¶§ulþÜ¡S§¦SOz¶ØKâd÷iúª§k6þadl=n·rotadS=0lS22(2)Ïl:x+y=4,z=1§dlÐ7Oz¶^=±§~êc>0¿©(c<2)§¦lu²222¡z=cþ§3²¡z=cþ7Oz¶±lr:x+y=r,z=c§r¿©§¦ru2§±llr>.Üþ¡SI§¦SIOz¶ØZZd÷iúª§a·dr+a·dr=n·rotadS=0§Ù¥−lrL«÷^l−lrIISu´6þa·dr=a·drllrIZ2πqlrëê§x=rcosθ,y=rsinθ,z=c§a·dr=dθ=2πIlr0la·dr=2π.l若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn2838.¦þa=−yi+xj+ck(c~ê)6þµ22(1)÷±x+y=1,z=0¶22(2)÷±(x−2)+y=1,z=0.)µ22(1)Ïl:x+y=1,z=0§Kl=costi+sintj+0k(06t62π)IZ2πu´a·dl=dt§l¤¦6þa·dl=dt=2πl022(2)Ïl:(x−2)+y=1,z=0§Kl=(2+cost)i+sintj+0k(06t62π)IZ2πu´a·dl=(2cost+1)dt§l¤¦6þa·dl=(2cost+1)dt=2πl09.y²µ(1)rot(uA)=urotA+gradu×A¶(2)div(φa)=φdiva+gradφ·a(3)graddiva−rotrota=∆ay²µ!!∂Az∂Ay∂u∂u(1)Ïrotx(uA)=u−+Az−Ay=urotxA+(gradu×A)x∂y∂z∂y∂zÓ{§roty(uA)=urotyA+(gradu×A)y,rotz(uA)=urotzA+(gradu×A)zu´rot(uA)=urotA+gradu×A∂(φax)∂ax∂φ∂(φay)∂ay∂φ∂(φaz)∂az∂φ(2)Ï=φ+ax,=φ+ay,=φ+az∂x∂x∂x∂y∂y∂y∂z∂z∂zKdiv(φa)=φdiva+gradφ·a!∂ax∂ay∂az(3)Ïgraddiva=grad++∂x∂y∂z!!!222222222∂ax∂ay∂az∂ax∂ay∂az∂ax∂ay∂az=++i+++j+++k∂x2∂x∂y∂x∂z∂x∂y∂y2∂y∂z∂x∂z∂y∂z∂z2"!!!#∂az∂ay∂ax∂az∂ay∂axrotrota=rot−i+−j+−k∂y∂z∂z∂x∂x∂y课后答案网!!!222222222222∂ay∂az∂ax∂ax∂ax∂az∂ay∂ay∂ax∂ay∂az∂az=+−−i++−−j++−−k∂x∂y∂x∂z∂y2∂z2∂x∂y∂y∂z∂x2∂z2∂x∂z∂y∂z∂x2∂y2Kgraddiva−rotrota=∆axi+∆ayj+∆azk=∆a10.¦rot(w×r)§w~¥þ§r¥»þ.)µw=(w1,w2,w3),r=(x,y,z)ijkwww.hackshp.cnu´w×r=w1w2w3=(w2z−w3y)i+(w3x−w1z)j+(w1y−w2x)kxyzrot(w×r)=2w1i+2w2j+2w2k=2w11.y²a=yz(2x+y+z)i+xz(x+2y+z)j+xy(x+y+2z)kÅ|§¿¦Ù³¼ê.y²µém?:(x,y,z)§k()∂∂rota=[xy(x+y+2z)]−[xz(x+2y+z)]i∂y∂z()∂∂+[yz(2x+y+z)]−[xy(x+y+2z)]j∂z∂x()∂∂+[xz(x+2y+z)]−[yz(2x+y+z)]k∂x∂y=0KaÅ|du³φ÷vdφ=a·dl=axdx+aydy+azdz=d[(xyz(x+y+z)]KÙ³¼êu(x,y,z)=xyz(x+y+z)+C§Ù¥C?¿~ê.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn28412.¦þa=r÷Úr=acosti+asintj+btk(06t62π)ã¤õ.)µÏa=xi+yj+Zzk,l:x=acost,y=asint,z=bt(062π)22K¤¦õW=xdx+ydy+zdz=2bπl13.φ¼ê§Oµgradφ(r),div(φ(r)r)9rot(φ(r)r).r00)µgradφ(r)=φ(r)gradr=φ(r)·r0div(φ(r)r)=φ(r)divr+r·gradφ(r)=3φ(r)+rφ(r)rot(φ(r)r)=φ(r)rotr+gradφ(r)×r=014.¦÷v^div(φ(r)r)=0¼êφ(r).0)µdþK§div(φ(r)r)=3φ(r)+rφ(r)0φ(r)30¦div(φ(r)r)=0§3φ(r)+rφ(r)=0==−φ(r)rcKφ(r)=(c~ê)r315.¦±eþÑÝ9^Ý(a,b~þ)µ(1)(a·r)b(2)a×r(3)φ(r)(a×r)(4)r×(a×r))µ(1)div[(a·r)b]=(a·r)divb+grad(a·r)·b=grad(a·r)·b=a·brot[(a·r)b]=(a·r)rotb+grad(a·r)×b=grad(a·r)×b=a×b(2)Ï(a×r)x=ayz−azy,(a×r)y=azx−axz,(a×r)z=axy−ayx∂∂∂Kdiv(a×r)=(ayz−azy)+(azx−axz)+(axy−ayx)=0∂x∂y∂zijk∂∂∂rot(a×r)==2a∂x∂y∂zayz−azyazx−axzaxy−ayx0(3)div[φ(r)(a×r)]=φ(r)div(a×r)+grad(φ(r))(a×r)=φ(r)(r·rota−a·rotr)+(rφ(r)·(a×r)=0rφ(r)0φ(r)·(a×r)=[r·(课后答案网a×r)]=0rrrot[φ(r)(a×r)]=φ(r)rot(a×r)+grad(φ(r))×(a×r)=!0rφ(r)02φ(r)[−(axi+ayj+azk)+3a]+φ(r)×(a×r)=2φ(r)a+[ra−(a·r)r]rr2(4)Ïr×(a×r)=|r|a−(a·r)r22Kdiv[r×(a×r)]=|r|diva+gradwww.hackshp.cn|r|·a−[(r·a)divr+r·grad(a·r)]=(ax+ay+az)−4(xax+yay+zaz)122rot[r×(a×r)]=|r|rotr+grad|r|×a−(r·a)rotr−grad(r·a)×r=(r×a)−a×r.r16.y²±e ªµ(1)grad(a·b)=a×rotb+b×rota+(b·∇)a+(a·∇)b(2)rot(a×b)=(b·∇)a−(a·∇)b+(divb)a−(diva)b(3)c·grad(a·b)=a·(c·∇)+b·(c·∇)a(4)(c·∇)(a×b)=a×(c·∇)b−b×(c·∇)ay²µ(1)grad(a·b)=grad(axbx+ayby+azbz)=!∂bx∂ax∂by∂ay∂bz∂azax+bx+ay+by+az+bzi+∂x∂x∂x∂x∂x∂x!∂bx∂ax∂by∂ay∂bz∂azax+bx+ay+by+az+bzj+∂y∂y∂y∂y∂y∂y若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn
课后答案网：www.hackshp.cn285!∂bx∂ax∂by∂ay∂bz∂azax+bx+ay+by+az+bzk∂z∂z∂z∂z∂z∂z=a×rotb+b×rota+(b·∇)a+(a·∇)b!!∂∂∂∂∂∂(2)rot(a×b)=bx+by+bza−ax+ay+azb+(divb)a−(diva)b∂x∂y∂z∂x∂y∂z=(b·∇)a−(a·∇)b+(divb)a−(diva)b(3)c·grad(a·b)=!∂bx∂ax∂by∂ay∂bz∂azcxax+bx+ay+by+az+bz+∂x∂x∂x∂x∂x∂x!∂bx∂ax∂by∂ay∂bz∂azcyax+bx+ay+by+az+bz+∂y∂y∂y∂y∂y∂y!∂bx∂ax∂by∂ay∂bz∂azczax+bx+ay+by+az+bz∂z∂z∂z∂z∂z∂z=a·(c·∇)+b·(c·∇)a!∂∂∂(4)(c·∇)(a×b)=cx+cy+cz[(aybz−azby)i+(azbx−axbz)j+(axby−aybx)k]∂x∂y∂z∂ay∂bz∂az∂by∂ay∂bz∂az∂by=bzcx+aycx−bycx−azcx+bzcy+aycy−bycy−azcy+∂x∂x∂x∂x∂y∂y∂y∂y!∂ay∂bz∂az∂bybzcz+aycz−bycz−azczi+∂z∂z∂z∂z∂az∂bx∂ax∂bz∂az∂bx∂ax∂bzbxcx+azcx−bzcx−axcx+bxcy+azcy−bzcy−axcy+∂x∂x∂x∂x∂y∂y∂y∂y!∂az∂bx∂ax∂bzbxcz+azcz−bzcz−axczj+∂z∂z∂z∂z∂ax∂by∂ay∂bx∂ax∂by∂ay∂bxbycx+axcx−bxcx−aycx+bycy+axcy−bxcy−aycy+∂x∂x∂x∂x∂y∂y∂y∂y!∂ax∂by∂ay∂bxbycz+axcz−bxcz−ayczk∂z∂z∂z∂z=a×(c·∇)b−b×(c·∇)a17.Áydivgradsin2rL«¤F(r)/ª§¿Ñ课后答案网F(x).222222∂sinr∂sinr∂sinr22y²µdivgradsinr=∆sinr=++=!!∂x2∂y2!∂z2∂x∂y∂z12sin2r·+sin2r·+sin2r·=2cos2r+2sin2r·=(sin2r+rcos2r)∂xr∂yr∂zrrr2=F(r)=(sin2r+rcos2r)www.hackshp.cnr22dsinx2KF(x)=divgradsinxr==2cos2xdx2218.y²µ|a|=~ê§k(a·∇)a=−a×rota.22y²µÏ|a|=~ê§Kgrad|a|=0d16K(1)§grad(a·a)=2[a×rota+(a·∇)a]=0K(a·∇)a=−a×rota.若侵犯了您的版权利益，敬请来信告知！www.hackshp.cn'
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