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1、高等数学上册答案第一章 函数与极限 更多答案 参见百度号 利万物而Bu争 1. 答:(1,10) (由得) 1,x,100,lgx,122. 答:(,2,2) (由0?x,4,解得,2,x,2) 3. ,,0 , 1(由arcsinx,0得0,x,1)24. (由0,1,x,1得x,1 ,x,0)(,1 , 0):(0 , 1)5. 0,1 6. (由x,x,0即x,x,?x,0)(, , 0)7. 答:(1,+,) (logx,0得x,1)222,x28. x,2且x,0由有2,x,0且x,0得x22 由ln(x,x)有x,x,0得x,0或x,1故函数的定义域为 ,,,2 , 0:1 , 2
2、2,x2,x9. 由ln,有,0解得,2,x,2;2,x2,x111,当x,0时,对f有x,或x, ,x22,111,故函数f(x),f的定义域为 (,2 , ,):( , 2),x22,2x2x1由arccos,有,1,解得,x,1;10. 1,x1,x3122 由1,x,2x,有1,x,2x,0,解得,1,x,;211,故函数的定义域为 , , ,32,由a,x,m,b且a,x,m,b11. a,m,x,b,m,得 ,a,m,x,b,m(m,0),b,a 当m,时,F(x) 的定义域为,;2b,a ,当0,m,时,F(x)的定义域为 a,m ,b,m22,x,22,x,0,12. 由得,1
3、,x,0x,1,故的定义域为,fx(),211112:,,22x,5xx,5x解得:或xx,6113. 由lg,0得,166故的定义域是,fx(),,,61U ,,2x,5,x250由得14. ,x,0x,0,故的定义域为,fx(),5005: ,,215. 由y|,x得f(x,1),x,1,(x,1),2(x,1)a,02 故fxxx(),,2这时yaxax,,,,11由得ya|,21故yx,,1x,1112当x,2 时,f(t,2),t,2t,516. 222令t,2,u,则t,u,2 ?f(t,2),t,4t,10222,u,6 f(u),(u,2),4(u,2),10?f(x),x,6
4、2217. 因 y,0 时 , z,x ?x,f(x),x22故有 f(x),x,xf(x,y),(x,y),(x,y)22?z,x,y,(x,y),(x,y),2y,(x,y) 21x,2x218. 已知:2f(x),xf(),(1)xx,112,2x112x,1x 故得:2f(),f(x),2x1x(x,1)x,1x212x,x2 f(x),2xf(),(2)xx,12212x,4x,2x,x3x2,(1),(2)消去f()得: 3f(x), xx,1x,1x故f(x), x,119. A 20.C 21. D 22. A 23.(1)取x,y,0时,有f(0),f(0),f(0)故f()
5、00,(2)取y,x,有f(0),f(x),f(,x) 于是f(x),f(,x),0即f(,x),f(x),x,(,,,)。 因此是奇函数fx()xx,2x1ee,e,24. 由得y, xx,2x1ee,e,1,y11,y2x e,x,ln1,y21,y11,x反函数,()lnx, 定义域,(),1121,x25 因,且xfxfx,001()()1,1fx()fx()f,1x ffx(),x,1fx()fx(),1,,1fx()从而xfffxfx,(),,x,126. 定义域,;,12值域,。02,,42,,xx,;27. ffx(),22,x,28.D 29.C 30.D 31.C 32.A
6、 33.D 34.C 35.A 36.B 37.C 38.C 39.B 40.C cosxcosx,1(cos)ex,141.因当x,0,e,e,e(e,1)e2x,ee22 故原式,lim,x2x,022x212xxx,,sincos42. 原式,limx,0xxxxxtan(sincos)12,21xx2xsinsin15 ,,lim(),,,()14x,02xxtanxxtan2232,xaxxb43.因lim,32x,1,1x即ba,2 32x,ax,x,ba,b,2故lim(,1),0x2x,12,1x323xaxxb,x,x,11则limlim(),,a 222xx,11x,x,x
7、,11131?,ab13,,,,,,aa23 2244. B 12n设x,,?45. n222nnnn,1,2nnn,121,?n?12,nn,1则x,x, 又 nn222nn,22()n,nn,21n,11故limx,又lim, , nn,n,222()n,246. 0 nn,12x1x1x1x1, 47. 原式,lim(),?,x,1x,1x,1x,1x,1nn(),1 ,,,,nn()121?212648. 49. e5x,x,2xe2e12e1,50.limlim ,x,x,2xx,,,x,,,33e4xe34e,xx,2xee,2e221, 而lim,lim,22xx,xx,x,42
8、34e34ee,x,x11e,2e,lim因故原极限不存在 x,xx,323e,4e51.C 52.C 2333xxxxxx53.2203440ln(12)2当,,,,,,,32323arcsin3x4x43x4x4(x2)(3x2),,3x211,limlim原式,3x,2x,232(x2)(3x2)3x2,,xxx设u,coscoscos?54. n2n222xxxx11,coscoscossin?,sinx nn2xx2222nsinsin2nn22xnsinsinxxxsin2,limlimu,lim,1limlimu,n,则有 nx,0n,x,0n,nxxxxsinn2xxxxxxx
9、coscos(cos)1,55. eeee,1,xxxxxcoscos 当,xeeexx,01(cos)22 ln()1,xx1xx(cos),1xxcos 原式,lime,2x,02xx,56.D 57.A 11n(),12e2n,e,58. 原式,lim 1n,1,n2(),1e2n12x59. 原式,,lim(sin)12x x,012sinx,e22sinxx,,lim(sin)12x ,0x32x,12()1,,2x原式,lim60. 3,x()(),2x12()1,2x32e3,e ,3,2eln(sin),lncos1xx,61. 原式,lim,x,0sinsinxx,2ln(s
10、in),ln(sin),sin11xxx,lim,lim, 2xx,00sinx,sin2x,1,0,1 2xxee,,21原式,lim62. x2x,0,exxe,112,lim() xx,0xe,1 tanxx3e11e,63. 原式,lim(,)x,0sinsinxxtanxx3e11xe3x,tan, =1, ,lim,lim(,)x,00xxsinx3xsinxnxlne1,原式,lim64. ,1x,x1nxlnln(),,11x,e1,lim,n ,1xln,nxx1,n m,x1x1n,mn,x1x1,原式,lim65. x,1,x1x1mn,mn,,x1x1,66.因为g(,
11、0),lim(x,),x,022 ,g,,x,,(0)lim()x,,022于是不存在lim()gx x,0,sin()20xx,,,当而fgx(), sin()20xx,,当,sin2x所以lim()lim(sin)fgxx,20, xx,0011167.证s,,?,n222(n,1)(n,2)(2n)1111s,,?,, n222nnnn1111,,?,,sn2224n(2n)(2n)(2n)11即有,s n4nn11而lim,0,lim,0n,n,4nn,111? 因此lim,,222n,(n,1)(n,2)(2n),,0xxx33tan(tan)(tan),,68. 原式,lim,x,
12、sin(),x33tanx,3 ,,,limtan(tan)limxx3,xx,sin(),x333,xsin(),(),13 ,6lim,6,11,x,coscossin()xx,33322,24111169.显然:x,1,?,1n222223n(n,1) 1,x,,xnn2(n,1)即数列单调增x,n 111又x,,1?n22223n111,,1,?12,23,()nn,1 11111,,,,,,11()()()?,223nn,11,22n即数列有上界为x2,n 根据准则:单调有界数列还有极限因此:存在limxnn,70.当n,1时,x,2,21设当n,k时,x,2成立k 则当n,k,1时
13、,x,2,x,2,2,2k,1k由数学归纳法原理知:数列有上界为x2,n下面再用数学归纳法证明数列为单调增x,n当时,nxxx,,,,,12222211设时,成立nkxx,kk,1 则当时,nk,,1xxxxx,,,,,22kkkkk,11211所以数列单调增x,n根据单调有界数列必有极限,知存在limx nx,可设,则limxAA,20nn, 由,得limlimxxAA,,,,22n,n1n,n,解得:,舍去),因此AAx,212(lim 12nn,1,xsinx,cosx原式,lim71. x,0xtanx(1,sinx,cosx)1131xsinx1,cosx,lim(,),(1,),
14、x,02xtanxxtanx224sincossincosxx,原式,lim72. x,()coscos,xxx,x,1sin(),lim, x,xxx,coscos,12,sec 2cos,(tan)(sin),,,11xx原式,lim73. 3x,0(tansin),xxx111sin(cos)xx1,lim 3x,02x1sincosx1,x,lim 2x,02xx1, 4()()axx,,121 74. fx(),()()axxa,,121x, ()lim()lim11当时,afx,xx,11x,1211x,1()lim()lim2fx,xx,11 xaa,1,2得a,1()lim()
15、()31210axx,,1x,21故欲使,必须lim()lim()fxxaa,00112xx,221即a, 21121()()xx,,2lim()limfx,2 1111xx,()()xx,,1222232xx,5x23ee,23,e1limlim,75. xx,x325x,,,x,,,24ee,4,e3x,2x5xeee2,32,3而lim,lim,3 325x,xxx,x,eee4,4,13x,2x2e,3e因此lim不存在. 3x,2xx,e,e41,,cos2x2x,76. 原式,lim x,13,sin3xx2,3n1077. 原式,lim,15n,32,1n,101012nnn,3
16、78. 原式,lim3 (),(),1 ,n33111k,k,由()1,79. 2kkk1324n11n,,原式,lim()()()?, n,2233nn1n1,,lim n,2n1, 2n80.当a,1时,a所以lim,0 nnn,2,a因为lima,0n,当时,a,1na1lim,lim,1所以 1nnn,n,1n2,a因为lim(),02(),1n,aa2,1nnn1(),原式,lim,81. ,n,n222,1,n,lim n,2(),22n82.B 83.B 84.C 85.C 86.A 87.D 88.D 89.D 90.B 91.B 92.A 93.C 94.C 95.B 96.
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