一元函数微积分复习题1-1到1-9答案

1）y1x29；解：要使函数有意义，则：x290即x3或x3.所以函数定义域：感谢阅读(,).）ylogarcsinx；a解：要使函数有意义，则arcsinx00x1）y1感谢阅读1xx2；1解：1x0且x101x1且x1感谢阅读2）y31x2a(2x；33

x20且2x30x2且x(,2)(2,).

22谢谢阅读x1）yarccos(4x2)；精品文档放心下载2ax111且4x201x3且2x2[精品文档放心下载21

）y.

sinx_sinx0则xk,kZZZ或xxk,kZ.谢谢阅读1sin,x02yx和f(0).谢谢阅读f0x0(,).11当x0,0故11.谢谢阅读xx2f()sin1,f(0)0.

2感谢阅读f(x)和g(x)?精品文档放心下载f(x)x,g(x)x2;精品文档放心下载专业专注.:g(x)x2|x|,即g(x)而f(x).谢谢阅读f(x)cosx,g(x)12sin2x2;:f(x)和g(x)的定义域均为实数精品文档放心下载因为值域为且g(x)12sin2x2cosxf(x)f(x)x21,g(x)x1;x1:f(x)xx21.x1x,g(x)xf(x)

0.x:f(x)x0),g(x)x0xxx设f(x)sinx,:f(xx)f(x)cos(x).2sin22xx

::f(xx)f(x)sin(xx)sinx2sincos(x).

22谢谢阅读设f(x)axbx5且f(xf(x)8x3,b.谢谢阅读2:f(x)axbx52故f(xa(xb(x5ax(2ab)x(ab精品文档放心下载22f(xf(x)2axa58x3精品文档放心下载:2a8且ab3:a1.??谢谢阅读yxx);22:(,)专业专注.f(x)(x)2(x)2]x2x2)f(x)感谢阅读.y3xx;23:(,)f(x)3(x)2(x)33x2x3,f(x)f(x)且f(x)f(x).谢谢阅读.1x2y;1x2:(,)1(x)21x2f(x)f(x)1(x)1x22.yx(x:(,)f(x)x(xx3x,f(x)(x)3(x)x3xf(x).精品文档放心下载.ysinxcosx1;:(,)f(x)sin(x)cos(x)1sinxx1则f(x)f(x)且f(x)f(x)谢谢阅读.axax

y.

2:(,)axaxf(x)f(x)

2.设f(x)(,):感谢阅读专业专注.(x)f(x)f(x);(x)f(x)f(x)F.F12:f(x)(,),则F(x),(x)F(,)12F(x)f(x)f(x)(x)f(x)f(x)(x)故F(x).FF1111F(x)f(x)f(x)(x)f(x)f(x)(x),.故F(x).FF2222:(,).谢谢阅读:设f(x)(,).精品文档放心下载(x)f(x)f(x)(x)f(x)f(x)F.由7题知F1211(x).且f(x)F(x)

F.2212设f(x)(L,L)若f(x)在(0,L),:f(x)在(谢谢阅读.:x(L,L):f(x)f(x)精品文档放心下载x,xLx0,则L122xx1f(x)在(0,L)上单增,则f)f(x(x1f(x)f),f(x)f)(x(x2112即f(x)f)(x12f(x)f)(x1222x0.1)，f(x)奇函数f(x)在(.?:感谢阅读y;cos(x22)2),T2.谢谢阅读ycos;专业专注.:cos4(x2)cos(4x2)cos,期T2.y1sinx;:1sinx1x2)1sin(xT2.精品文档放心下载yxcosx;:ysinx;2解:sin2x111cos2x)cos(2x2)]cos2(x)],函222T.ysintanx2解:sinsin(3x2)sin),tanx)谢谢阅读32数sin3x,tanx和.所以周期为T2.谢谢阅读3?,谢谢阅读.yx,x;3:ysint,:(,).精品文档放心下载3ya,ux;u

2:yax,:(,).感谢阅读2yu,u3xloga2;2:yloga(2x2),:(,).2yu,usinx2;:yuu0,usinx2[.谢谢阅读yu,ux3;:yx3,:[0,).谢谢阅读ylogau,ux2.2专业专注.:y(x2),:(,2)(2,).2

a?y3x)21;:y3u,ux)1.谢谢阅读2y3(ln2x;:y3,uv,vx1.谢谢阅读u2ysin(3x;3:yu,usinv,v3x1.感谢阅读3y3cosx.2

a:y3u,ulogv,vw2,wcosx.感谢阅读a:y2sinx;x[,]

22谢谢阅读y

::x[,],[2,2].:x.

222谢谢阅读x:y.2arcsiny1log(x2);a::(),(,).:xay12.:ya2x1(,):().精品文档放心下载y22x1x.:y2x2x11111211,则01.x21212121xxxxyy:(,),(:2,x1yx1y2.专业专注.x:ylog12x:,:(,).谢谢阅读:感谢阅读202.50;谢谢阅读50202.30;

10050;精品文档放心下载感谢阅读100100;谢谢阅读设x,y,yx.感谢阅读::当0x,y2.5x;谢谢阅读当20x,y2.52.3x4;精品文档放心下载当50x,y2.52.32(x2x;精品文档放心下载当x,y1.8x精品文档放心下载2.5x0x

202.3x4

x精品文档放心下载y2xx1.8xx)QS(p)4p,Q为供应量,p为市场价格.谢谢阅读,Lp.精品文档放心下载:QS(p)4p谢谢阅读L(p1.5)Q(p1.5)(804p)4p286p120.精品文档放心下载用p,某商品的需求函数为QD(p)750p,当Q1时感谢阅读C,(提示:当总收入=精品文档放心下载:当Q时QD(p)7p.感谢阅读,则C:p(750p)50p)精品文档放心下载专业专注.p2165p39000得p107.822p,故p28.59:28.59.r,Vh,

.感谢阅读谢谢阅读hh2:Rr()r感谢阅读222224:VRh(r22h21)hr2hh3.44:(0,2r)F,在,32F或212F或感谢阅读谢谢阅读F;

;?:F,,yF,精品文档放心下载yba,b.感谢阅读:a0ba1.8212100abb32:y1.8xxy感谢阅读),12精品文档放心下载011111n。nn111,nN11感谢阅读nn1)1。n专业专注.1(n24200。n2n2n2n谢谢阅读21(n1,nN0n2精品文档放心下载1(nn2n0。0x0x，精品文档放心下载•xU)x0x精品文档放心下载x0nx2404x2，x2感谢阅读x24•xU(2,)4x2x2感谢阅读x244nx2111100(x即x使1x1x1x1X11xX有1nx101x100(exsinx0exsinxexx感谢阅读Xln0xXexsinx0精品文档放心下载limesinx0xnlimf(x)x12x1x1f(x)lim(x2x1x1limf(x)f(x)感谢阅读x-x1专业专注.当x1f(x)x0xxxx0x1x0xx11xxxxx0x0x122x0xx11x2xx0x2xx0x1）sinxsinx感谢阅读x）limarctanx,arctanx感谢阅读22xx11）coscosxxx0）eex,limexxxxx）x1xx11x1xx11x1xx11limen0n1311）nn2n2在n小(n(n在nnn1n12n112n1,nn2n在n小1cosn1(n2）nnn01cosn1cosn

lim0，在nnnn11x0，xsinx,精品文档放心下载xx211x,,,ee精品文档放心下载xxxx2专业专注.）1xx小11

(x0xsinxxsin,0xx1当0x0xxsin、M(M,)xcosx100cosxMyxcosx在(,)yxx000MX总(X,)0y有xcosxx00yxcosxx精品文档放心下载0cosx0M01exx0x01e0x习题、求下列极限（）11111111解：lim(=)()(谢谢阅读)n1223n(n223nn1n感谢阅读1=)1nn112n12n（）解：lim()=nnnnnn22221

1n(nn1n=1

===n2n2n2nn22x25225

（）解：=x2x323x22x1(x2x10（）解：==0x1x1x1(xx1x122精品文档放心下载(xh)2x2(xhx)(xhx)感谢阅读（）解：==lim(2xh)2xh0hh0hh0谢谢阅读（）解：limx13xx11=x1(3x3x23xx(xx3x23x专业专注.=x1(xx(x3x23x=x1(x(3x23x=23x2525

（）解：==3

x1x313x111x3（）解：因为==0=x1x1x13x1x23(3)230（）解：===0x3x1(3)142

21

1x3x=1x2=（）解：31

22xx2x3x2xx

13

311

0xxxx02

2（）解：==x1311xx3x213xx313xx2)3(x感谢阅读（）解：()==x11x1x1x1x精品文档放心下载333

x1x1(x==1x1xx、求下列极根1210001000n0

（）解：n=

nn01=x1112n2n2nlim11n（）解：=n2nn21n1an11aa2an1a=1b（）解：=n1bb2bn1bn11an1b121()n33(2)n3n（）解：lim=1n(n1=3(2)n33n1感谢阅读3n1n(2)3专业专注.x

）x1x14x21(2x2x1感谢阅读=lim==416x25x11(2x13x1xxx222精品文档放心下载、求下列极限：（）解：(exsinx)=xx11x0x为无穷小量，cos为有界量）（）解：xcos0时，（当x0xx（）解：sinn0（当n0为无穷小量，sin谢谢阅读n为有界量）nnn（）解：0xxex（）解：因lime，arctanx，所以谢谢阅读xxx2x（）解：xarccotx0ex、下列各题的做法是否正确？为什么？（）错。因为分式分母的极限为零，故不能利用极限的商的运算。谢谢阅读x90x9正确做法：因0，所以x9x9x9

22

x911（2）错。因为差式中与都不存在，故不能利用差的运算求极精品文档放心下载x1x1x1x12限。11(x11正确做法：lim()==，x1x1x1x1x21x1x212感谢阅读x210

因=0

x1x1所以x111)x1x12（）错误。因为limcosx不存在，故不能利用乘积的根限运算。精品文档放心下载x正确做法：1因当xcosx为有界量，利用无穷小量与有界量乘感谢阅读x积是无穷小量。cos

x0所以xx5专业专注.sinasinax0sinx0bb谢谢阅读tanxsinxcosx)tanx2sin2x2x2x0x3xx3x0xsinx21x121精品文档放心下载2x0xcosxx221cosxx0xsinx2sin2xx0x2xsincos22x2xsin1212xx02xcos1222xtanx2xtanxx1感谢阅读2211x0sinxx0sinxxxx0sinxx0x精品文档放心下载谢谢阅读arcsinxxxx精品文档放心下载xyy1x0sinxx0xxx0xy0y感谢阅读精品文档放心下载33xx333xlim1lim11e333xxxxxx谢谢阅读精品文档放心下载11111xlim1lim1xxxxxxlim1xxe1lim1xx1x21111x2x2lim11lim1lim1exxxxxxx11xytanx)cotxtanx)y)exyx0x0y0谢谢阅读xax2ax2axa2a感谢阅读alimlim111xxaxaxaxa谢谢阅读xx2a2axaxa2a2a2alim1alim1lim12axxa2axxxaxae2alimx22x211x21lim1exx21xx12专业专注.111）)))ene112xnnnnnn12n12n

xnnnnnnnnn222n2221n(n1n(n

2x2

nnn2n21n(n11n(n1又lim2，lim2xnnxnn222212n1xn2n2nn222谢谢阅读xn1xn2(n),xxn10(n)。2当nx22，nk1,x2，k当nk1x222x2，k1kx2(n)，n

n则x2xx(x2)(x又xx2x2n1nxnnnnn2x2xxnxnnnn由x2知xx0，则xnn1nn22,222,感谢阅读xa2a2axxxalimlim11xxaxxaxxa12aaxa2a2axa2a2a2axalim12axxaalim1lim12axxxaxae2a2a4aln2。

e5pk5n120n0.05lim101114112.844xnn20n专业专注.x~xx~x感谢阅读即1xx1xxxxxxxxxx即x~x。sin2x2x2

；

x0sinx03sin2x2xx0arctanxx0x1,nmsinxn

xnx0sin0,nxm；mx0xm,nmxx01xx0x12x22xsinx．1x0x2当x0x的2精品文档放心下载132sinxsinxcosxcos2x223．limx0x2x0x22

当x0是x的2b;a感谢阅读谢谢阅读fxx精品文档放心下载,00,；xf101f101f1；精品文档放心下载f101f101-1感谢阅读,1；x11limfxx0xx0xk2(k2,)专业专注.x而lim0xk(k2,)谢谢阅读；tanx2xk

2x当，ytan

xx而1谢谢阅读；x0tanxx当xk(k2,)y感谢阅读tanxx而xk(k2,)精品文档放心下载xktanx4f10ef10A1f11A精品文档放心下载故f001f001limfx1x0但在是8感谢阅读精品文档放心下载．f(x)在[a,b]limf(x)f(x),(x[a,b])f(x)00xx01111f(x)0(x[a,b])则在[a,b]0fxxx0f(x)f(x)f(x)0xx01x22x55；）lim(sin2x)31x0x42cosx