一元函數(shù)微積分復習題1-1到1-9答案

1）y1x29；解：要使函數(shù)有意義，則：x290即x3或x3.所以函數(shù)定義域：感謝閱讀(,).）ylogarcsinx；a解：要使函數(shù)有意義，則arcsinx00x1）y1感謝閱讀1xx2；1解：1x0且x101x1且x1感謝閱讀2）y31x2a(2x；33

x20且2x30x2且x(,2)(2,).

22謝謝閱讀x1）yarccos(4x2)；精品文檔放心下載2ax111且4x201x3且2x2[精品文檔放心下載21

）y.

sinx_sinx0則xk,kZZZ或xxk,kZ.謝謝閱讀1sin,x02yx和f(0).謝謝閱讀f0x0(,).11當x0,0故11.謝謝閱讀xx2f()sin1,f(0)0.

2感謝閱讀f(x)和g(x)?精品文檔放心下載f(x)x,g(x)x2;精品文檔放心下載專業(yè)專注.:g(x)x2|x|,即g(x)而f(x).謝謝閱讀f(x)cosx,g(x)12sin2x2;:f(x)和g(x)的定義域均為實數(shù)精品文檔放心下載因為值域為且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設f(x)sinx,:f(xx)f(x)cos(x).2sin22xx

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

22謝謝閱讀設f(x)axbx5且f(xf(x)8x3,b.謝謝閱讀2:f(x)axbx52故f(xa(xb(x5ax(2ab)x(ab精品文檔放心下載22f(xf(x)2axa58x3精品文檔放心下載:2a8且ab3:a1.??謝謝閱讀yxx);22:(,)專業(yè)專注.f(x)(x)2(x)2]x2x2)f(x)感謝閱讀.y3xx;23:(,)f(x)3(x)2(x)33x2x3,f(x)f(x)且f(x)f(x).謝謝閱讀.1x2y;1x2:(,)1(x)21x2f(x)f(x)1(x)1x22.yx(x:(,)f(x)x(xx3x,f(x)(x)3(x)x3xf(x).精品文檔放心下載.ysinxcosx1;:(,)f(x)sin(x)cos(x)1sinxx1則f(x)f(x)且f(x)f(x)謝謝閱讀.axax

y.

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

2.設f(x)(,):感謝閱讀專業(yè)專注.(x)f(x)f(x);(x)f(x)f(x)F.F12:f(x)(,),則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:(,).謝謝閱讀:設f(x)(,).精品文檔放心下載(x)f(x)f(x)(x)f(x)f(x)F.由7題知F1211(x).且f(x)F(x)

F.2212設f(x)(L,L)若f(x)在(0,L),:f(x)在(謝謝閱讀.:x(L,L):f(x)f(x)精品文檔放心下載x,xLx0,則L122xx1f(x)在(0,L)上單增,則f)f(x(x1f(x)f),f(x)f)(x(x2112即f(x)f)(x12f(x)f)(x1222x0.1)，f(x)奇函數(shù)f(x)在(.?:感謝閱讀y;cos(x22)2),T2.謝謝閱讀ycos;專業(yè)專注.:cos4(x2)cos(4x2)cos,期T2.y1sinx;:1sinx1x2)1sin(xT2.精品文檔放心下載yxcosx;:ysinx;2解:sin2x111cos2x)cos(2x2)]cos2(x)],函222T.ysintanx2解:sinsin(3x2)sin),tanx)謝謝閱讀32數(shù)sin3x,tanx和.所以周期為T2.謝謝閱讀3?,謝謝閱讀.yx,x;3:ysint,:(,).精品文檔放心下載3ya,ux;u

2:yax,:(,).感謝閱讀2yu,u3xloga2;2:yloga(2x2),:(,).2yu,usinx2;:yuu0,usinx2[.謝謝閱讀yu,ux3;:yx3,:[0,).謝謝閱讀ylogau,ux2.2專業(yè)專注.:y(x2),:(,2)(2,).2

a?y3x)21;:y3u,ux)1.謝謝閱讀2y3(ln2x;:y3,uv,vx1.謝謝閱讀u2ysin(3x;3:yu,usinv,v3x1.感謝閱讀3y3cosx.2

a:y3u,ulogv,vw2,wcosx.感謝閱讀a:y2sinx;x[,]

22謝謝閱讀y

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

222謝謝閱讀x:y.2arcsiny1log(x2);a::(),(,).:xay12.:ya2x1(,):().精品文檔放心下載y22x1x.:y2x2x11111211,則01.x21212121xxxxyy:(,),(:2,x1yx1y2.專業(yè)專注.x:ylog12x:,:(,).謝謝閱讀:感謝閱讀202.50;謝謝閱讀50202.30;

10050;精品文檔放心下載感謝閱讀100100;謝謝閱讀設x,y,yx.感謝閱讀::當0x,y2.5x;謝謝閱讀當20x,y2.52.3x4;精品文檔放心下載當50x,y2.52.32(x2x;精品文檔放心下載當x,y1.8x精品文檔放心下載2.5x0x

202.3x4

x精品文檔放心下載y2xx1.8xx)QS(p)4p,Q為供應量,p為市場價格.謝謝閱讀,Lp.精品文檔放心下載:QS(p)4p謝謝閱讀L(p1.5)Q(p1.5)(804p)4p286p120.精品文檔放心下載用p,某商品的需求函數(shù)為QD(p)750p,當Q1時感謝閱讀C,(提示:當總收入=精品文檔放心下載:當Q時QD(p)7p.感謝閱讀,則C:p(750p)50p)精品文檔放心下載專業(yè)專注.p2165p39000得p107.822p,故p28.59:28.59.r,Vh,

.感謝閱讀謝謝閱讀hh2:Rr()r感謝閱讀222224:VRh(r22h21)hr2hh3.44:(0,2r)F,在,32F或212F或感謝閱讀謝謝閱讀F;

;?:F,,yF,精品文檔放心下載yba,b.感謝閱讀:a0ba1.8212100abb32:y1.8xxy感謝閱讀),12精品文檔放心下載011111n。nn111,nN11感謝閱讀nn1)1。n專業(yè)專注.1(n24200。n2n2n2n謝謝閱讀21(n1,nN0n2精品文檔放心下載1(nn2n0。0x0x，精品文檔放心下載?xU)x0x精品文檔放心下載x0nx2404x2，x2感謝閱讀x24?xU(2,)4x2x2感謝閱讀x244nx2111100(x即x使1x1x1x1X11xX有1nx101x100(exsinx0exsinxexx感謝閱讀Xln0xXexsinx0精品文檔放心下載limesinx0xnlimf(x)x12x1x1f(x)lim(x2x1x1limf(x)f(x)感謝閱讀x-x1專業(yè)專注.當x1f(x)x0xxxx0x1x0xx11xxxxx0x0x122x0xx11x2xx0x2xx0x1）sinxsinx感謝閱讀x）limarctanx,arctanx感謝閱讀22xx11）coscosxxx0）eex,limexxxxx）x1xx11x1xx11x1xx11limen0n1311）nn2n2在n小(n(n在nnn1n12n112n1,nn2n在n小1cosn1(n2）nnn01cosn1cosn

lim0，在nnnn11x0，xsinx,精品文檔放心下載xx211x,,,ee精品文檔放心下載xxxx2專業(yè)專注.）1xx小11

(x0xsinxxsin,0xx1當0x0xxsin、M(M,)xcosx100cosxMyxcosx在(,)yxx000MX總(X,)0y有xcosxx00yxcosxx精品文檔放心下載0cosx0M01exx0x01e0x習題、求下列極限（）11111111解：lim(=)()(謝謝閱讀)n1223n(n223nn1n感謝閱讀1=)1nn112n12n（）解：lim()=nnnnnn22221

1n(nn1n=1

===n2n2n2nn22x25225

（）解：=x2x323x22x1(x2x10（）解：==0x1x1x1(xx1x122精品文檔放心下載(xh)2x2(xhx)(xhx)感謝閱讀（）解：==lim(2xh)2xh0hh0hh0謝謝閱讀（）解：limx13xx11=x1(3x3x23xx(xx3x23x專業(yè)專注.=x1(xx(x3x23x=x1(x(3x23x=23x2525

（）解：==3

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

21

1x3x=1x2=（）解：31

22xx2x3x2xx

13

311

0xxxx02

2（）解：==x1311xx3x213xx313xx2)3(x感謝閱讀（）解：()==x11x1x1x1x精品文檔放心下載333

x1x1(x==1x1xx、求下列極根1210001000n0

（）解：n=

nn01=x1112n2n2nlim11n（）解：=n2nn21n1an11aa2an1a=1b（）解：=n1bb2bn1bn11an1b121()n33(2)n3n（）解：lim=1n(n1=3(2)n33n1感謝閱讀3n1n(2)3專業(yè)專注.x

）x1x14x21(2x2x1感謝閱讀=lim==416x25x11(2x13x1xxx222精品文檔放心下載、求下列極限：（）解：(exsinx)=xx11x0x為無窮小量，cos為有界量）（）解：xcos0時，（當x0xx（）解：sinn0（當n0為無窮小量，sin謝謝閱讀n為有界量）nnn（）解：0xxex（）解：因lime，arctanx，所以謝謝閱讀xxx2x（）解：xarccotx0ex、下列各題的做法是否正確？為什么？（）錯。因為分式分母的極限為零，故不能利用極限的商的運算。謝謝閱讀x90x9正確做法：因0，所以x9x9x9

22

x911（2）錯。因為差式中與都不存在，故不能利用差的運算求極精品文檔放心下載x1x1x1x12限。11(x11正確做法：lim()==，x1x1x1x1x21x1x212感謝閱讀x210

因=0

x1x1所以x111)x1x12（）錯誤。因為limcosx不存在，故不能利用乘積的根限運算。精品文檔放心下載x正確做法：1因當xcosx為有界量，利用無窮小量與有界量乘感謝閱讀x積是無窮小量。cos

x0所以xx5專業(yè)專注.sinasinax0sinx0bb謝謝閱讀tanxsinxcosx)tanx2sin2x2x2x0x3xx3x0xsinx21x121精品文檔放心下載2x0xcosxx221cosxx0xsinx2sin2xx0x2xsincos22x2xsin1212xx02xcos1222xtanx2xtanxx1感謝閱讀2211x0sinxx0sinxxxx0sinxx0x精品文檔放心下載謝謝閱讀arcsinxxxx精品文檔放心下載xyy1x0sinxx0xxx0xy0y感謝閱讀精品文檔放心下載33xx333xlim1lim11e333xxxxxx謝謝閱讀精品文檔放心下載11111xlim1lim1xxxxxxlim1xxe1lim1xx1x21111x2x2lim11lim1lim1exxxxxxx11xytanx)cotxtanx)y)exyx0x0y0謝謝閱讀xax2ax2axa2a感謝閱讀alimlim111xxaxaxaxa謝謝閱讀xx2a2axaxa2a2a2alim1alim1lim12axxa2axxxaxae2alimx22x211x21lim1exx21xx12專業(yè)專注.111）)))ene112xnnnnnn12n12n

xnnnnnnnnn222n2221n(n1n(n

2x2

nnn2n21n(n11n(n1又lim2，lim2xnnxnn222212n1xn2n2nn222謝謝閱讀xn1xn2(n),xxn10(n)。2當nx22，nk1,x2，k當nk1x222x2，k1kx2(n)，n

n則x2xx(x2)(x又xx2x2n1nxnnnnn2x2xxnxnnnn由x2知xx0，則xnn1nn22,222,感謝閱讀xa2a2axxxalimlim11xxaxxaxxa12aaxa2a2axa2a2a2axalim12axxaalim1lim12axxxaxae2a2a4aln2。

e5pk5n120n0.05lim101114112.844xnn20n專業(yè)專注.x~xx~x感謝閱讀即1xx1xxxxxxxxxx即x~x。sin2x2x2

；

x0sinx03sin2x2xx0arctanxx0x1,nmsinxn

xnx0sin0,nxm；mx0xm,nmxx01xx0x12x22xsinx．1x0x2當x0x的2精品文檔放心下載132sinxsinxcosxcos2x223．limx0x2x0x22

當x0是x的2b;a感謝閱讀謝謝閱讀fxx精品文檔放心下載,00,；xf101f101f1；精品文檔放心下載f101f101-1感謝閱讀,1；x11limfxx0xx0xk2(k2,)專業(yè)專注.x而lim0xk(k2,)謝謝閱讀；tanx2xk

2x當，ytan

xx而1謝謝閱讀；x0tanxx當xk(k2,)y感謝閱讀tanxx而xk(k2,)精品文檔放心下載xktanx4f10ef10A1f11A精品文檔放心下載故f001f001limfx1x0但在是8感謝閱讀精品文檔放心下載．f(x)在[a,b]limf(x)f(x),(x[a,b])f(x)00xx01111f(x)0(x[a,b])則在[a,b]0fxxx0f(x)f(x)f(x)0xx01x22x55；）lim(sin2x)31x0x42cosx