文登考研数学--高等数学--习题集及其答案

1第一章函数·极限·连续一.填空题1.已知,__________)(,1)]([,sin)(2xxxfxxf则定义域为___________.解.21)(sin)]([xxxf,)1arcsin()(2xx1112x,2||,202xx2．设ataxxdttexx1lim,则a=________.解.可得atadttee=aatteaeaete)(,所以a=2.3.nnnnnnnnn2222211lim=________.解.nnnnnnnnnn22221nnnnnnnn222221111211222nnnnnnn所以nnnn221nnnnnnnn22222111212nnn212)1(2122nnnnnnnnn,(n)2112)1(12122nnnnnnn,(n)所以nnnnnnnnn2222211lim=214.已知函数01)(xf1||1||xx,则f[f(x)]_______.解.f[f(x)]=1.5.)3(limnnnnn=_______.解.nnnnnnnnnnnnnnnnnn3)3)(3(lim)3(lim=233limnnnnnnnnn6.设当xbxaxexfxx为时11)(,0的3阶无穷小,则.___________,ba2解.3030301lim)1(1lim11limxaxbxeebxxaxbxeexbxaxekxxxxxxxx203limxabxebeexxxx(1)2062limxbxebeexxxx(2)由(1):01)(lim0ababxebeexxxx由(2):021)2(lim0bbxebeexxxx21,21ab7.xxxx1sin1cotlim0=______.解.616sinlim3cos1limsinlimsinsinsincoslim020300xxxxxxxxxxxxxxxxx8.已知Annnkkn)1(lim1990(0),则A=______,k=_______.解.Aknnnnnknkkn119901990lim)1(lim所以k－1=1990,k=1991;1991111kAAk，二.选择题1.设f(x)和(x)在(－,+)内有定义,f(x)为连续函数,且f(x)0,(x)有间断点,则(a)[f(x)]必有间断点(b)[(x)]2必有间断点(c)f[(x)]必有间断点(d))()(xfx必有间断点解.(a)反例01)(x1||1||xx,f(x)=1,则[f(x)]=1(b)反例11)(x1||1||xx,[(x)]2=1(c)反例01)(x1||1||xx,f(x)=1,则f[(x)]=1(d)反设g(x)=)()(xfx在(－,+)内连续,则(x)=g(x)f(x)在(－,+)内连续,矛盾.所以(d)是答案.32.设函数xexxxfsintan)(,则f(x)是(a)偶函数(b)无界函数(c)周期函数(d)单调函数解.(b)是答案.3.函数2)2)(1()2sin(||)(xxxxxxf在下列哪个区间内有界(a)(－1,0)(b)(0,1)(c)(1,2)(d)(2,3)解.42sin)0(,42sin)0(,)(lim,)(lim01ffxfxfxx所以在(－1,0)中有界,(a)为答案.4.当11211,1xexxx函数时的极限(a)等于2(b)等于0(c)为(d)不存在,但不为解.01001)1(lim11lim1111121xxexexxxxxx.(d)为答案.5.极限222222)1(12325213limnnnn的值是(a)0(b)1(c)2(d)不存在解.222222)1(12325213limnnnn=1)1(11lim)1(1131212111lim2222222nnnnn,所以(b)为答案.6.设8)1()1()1(lim502595xaxxx,则a的值为(a)1(b)2(c)58(d)均不对解.8=502595)1()1()1(limxaxxx=100502559595/)1(/)1(/)1(limxxxaxxxx=5502595)/11()/1()/11(limaxxaxx,58a,所以(c)为答案.7.设)23()5)(4)(3)(2)(1(limxxxxxxx,则,的数值为(a)=1,=31(b)=5,=31(c)=5,=531(d)均不对解.(c)为答案.8.设232)(xxxf,则当x0时(a)f(x)是x的等价无穷小(b)f(x)是x的同阶但非等价无穷小4(c)f(x)比x较低价无穷小(d)f(x)比x较高价无穷小解.xxxx232lim0=3ln2ln13ln32ln2lim0xxx,所以(b)为答案.9.设6)31)(21)(1(lim0xaxxxx,则a的值为(a)－1(b)1(c)2(d)3解.0)31)(21)(1(lim0axxxx,1+a=0,a=－1,所以(a)为答案.10.设02)1()21ln()cos1(tanlim2202caedxcxbxaxx，其中,则必有(a)b=4d(b)b=－4d(c)a=4c(d)a=－4c解.2=)1()21ln()cos1(tanlim20xxedxcxbxa=caxdexcxbxaxx22212sincoslim220,所以a=－4c,所以(d)为答案.三.计算题1.求下列极限(1)xxxex1)(lim解.eeeeeexxxxxxxexexexxexxxxx11lim)ln(lim)ln(1lim)(lim(2)xxxx)1cos2(sinlim解.令xy1yyxxyyxx10)cos2(sinlim)1cos2(sinlim=2cos2sinsin2cos2lim)cos2ln(sinlim00eeeyyyyyyyyy(3)310sin1tan1limxxxx解.310sin1tan1limxxxx310sin1sintan1limxxxxx3)sin1(sintansintansin10sin1sintan1limxxxxxxxxxxx=30sintanlimxxxxe=30)cos1(sinlimxxxxe=212sin2sinlim320eexxxx.2.求下列极限5(1)323112arcsin)11ln(limxxx解.当x1时,331~)11ln(xx,323212~12arcsinxx.按照等价无穷小代换33132313231221121lim121lim12arcsin)11ln(limxxxxxxxx(2)xxx220cot1lim解.方法1：xxx220cot1lim=xxxx2220sincos1lim=xxxxxx222220sincossinlim=4220cos)1(1limxxxx=32204sincos)1(2cos2limxxxxxxx=3203204sincos2lim42sincos2limxxxxxxxxxx=21122cos2sincos4cos2lim220xxxxxxx=2131242sin4sincos4lim2131122cos2cos2lim0220xxxxxxxxx=322131612131242sin2lim0xxx方法2：xxx220cot1lim=xxxx2220sincos1lim=xxxxxx222220sincossinlim=4220cos)1(1limxxxx=420)12)(cos1(211limxxxx=444220)(0!4)2(!2)2(11)(1(211limxxxxxx=4442420))(024162222(211limxxxxxxx6=3232lim440xxx3.求下列极限(1))1(lnlimnnnnn解.nnnnnnnnnnln1lim)1(lnlimxnn1令1)1ln(lim0xxx(2)nxnxnee11lim解.10111limnxnxnee000xxx(3)nnnnba2lim,其中a0,b0解.nnnnba2limabcnx/,/1xcxxxxxaeca2ln)1ln(lim10021lim=ababacaaeaexxxxxcccxc1lnlim2ln)1ln(lim004.设0cos1010)cos1(2)(022xdttxxxxxxfx试讨论)(xf在0x处的连续性与可导性.解.20200200coslim1cos1lim)0()(lim)0('xxdttxdttxxfxffxxxxx0221lim21coslim2020xxxxxx320200)cos1(2lim1)cos1(2lim)0()(lim)0('xxxxxxxfxffxxx06)1(cos2lim32sin2lim020xxxxxxx所以0)0('f,)(xf在0x处连续可导.5.求下列函数的间断点并判别类型7(1)1212)(11xxxf解.11212lim)0(110xxxf,11212lim)0(110xxxf所以x=0为第一类间断点.(2)11sincos2)2()(2xxxxxf00xx解.f(+0)=－sin1,f(－0)=0.所以x=0为第一类跳跃间断点;11sinlim)(lim211xxfxx不存在.所以x=1为第二类间断点;)2(f不存在,而2cos2)2(lim2xxxx,所以x=0为第一类可去间断点;xxxkxcos2)2(lim2,(k=1,2,…)所以x=2k为第二类无穷间断点.6.讨论函数xexxxf1sin)(00xx在x=0处的连续性.解.当0时)1sin(lim0xxx不存在,所