数学分析十讲习题册答案

1111/27272727习题1-11．计算下列极限（1）limxaxaaxxa→−−,0;a解：原式lim[]xaaaxaaaxaxaxa→−−=−−−=()|()|xaxaxaax==′′−=1lnaaaaaa−−⋅=(ln1)aaa−（2）sinsinlimsin()xaxaxa→−−；解：原式sinsinlimxaxaxa→−=−(sin)'cosxaxa===（3）21lim(2),0;nnnnaaa→∞+−解：原式211lim()1/nnnana→∞−=20[()']xxa==2lna=（4）1lim[(1)1]pnnn→∞+−，0;p解：原式111(1)1lim()|pppxnnnx=→∞+−′===11pxpxp−==（5）10100(1tan)(1sin)lim;sinxxxx→+−−解：原式101000(1tan)1(1sin)1limlimtansinxxxxxx→→+−−−=−−=990010(1)|10(1)|20tttt==+++=（6）11lim1mnxxx→−−，,mn为正整数；解：原式111lim11mnxxxxx→−−=−−i1111()'()'mxnxxx===nm=2．设()fx在0x处二阶可导，计算00020()2()()limhfxhfxfxhh→+−+−.解：原式000()()lim2hfxhfxhh→′′+−−=00000()()()()lim2hfxhfxfxfxhh→′′′′+−+−−=000000()()()()limlim22hhfxhfxfxhfxhh→→′′′′+−−−=+−00011()()()22fxfxfx′′′′′′=+=2222/272727273．设0a，()0fa，()fa′存在，计算1lnln()lim[]()xaxafxfa−→.解：1lnln()lim[]()xaxafxfa−→ln()ln()lnlnlimfxfaxaxae−−→=ln()ln()limlnlnxafxfaxae→−−=ln()ln()limlnlnxafxfaxaxaxae→−−−−=i'()()faafae=i习题1-21.求下列极限（1）lim(sin1sin1)xxx→+∞+−−;解：原式1limcos[(1)(1)]02xxxξξ→+∞=+−−=i，其中ξ在1x−与1x+之间（2）40cos(sin)coslimsinxxxx→−;解：原式=40sin(sin)limxxxxξ→−−=30sinsinlim()()()xxxxxξξξ→−−⋅=16，其中ξ在x与sinx之间（3）666565lim();xxxxx→+∞+−−解：原式116611lim[(1)(1)]xxxx→+∞=+−−56111lim(1)[(1)(1)]6xxxxξ−→+∞=⋅+⋅+−−5611lim(1)33xξ−→+∞=+=，其中ξ在11x−与11x+之间（4）211lim(arctanarctan);1nnnn→+∞−+解：原式22111lim()11nnnnξ→+∞=−++i1=，其中其中ξ在11n+与1n之间2．设()fx在a处可导，()0fa，计算11()lim()nnnnfafa→∞⎡⎤+⎢⎥−⎣⎦.解：原式1111(ln()ln())lim(ln()ln())limnnfafanfafannnnnee→∞+−−+−−→∞==11ln()ln()ln()ln()[limlim]11nnfafafafannnne→∞→∞+−−−+−=()()2()()()()fafafafafafaee′′′+==习题1-31-31-31-31111．求下列极限3333/27272727（1111）0(1)1lim(1)1xxxλµ→+−+−,,,,0;µ≠解：原式0limxxxλλµµ→==（2222）201coscos2coslim11xxxnxx→−⋅⋅⋅+−;;;;解：02lncoscos2coslim12xxxnxIx→−⋅⋅⋅=20lncoslncos2lncos2limxxxnxx→++⋅⋅⋅+=−20cos1cos21cos12limxxxnxx→−+−+⋅⋅⋅+−=−22220(2)()limxxxnxx→++⋅⋅⋅+=21nii==∑（3333）011lim)1xxxe→−−（;;;;解：原式01lim(1)xxxexxe→−−=−201limxxexx→−−=01lim2xxex→−=01lim22xxx→==（4444）112lim[(1)]xxxxxx→+∞+−;解：原式11ln(1)ln2lim()xxxxxxee+→+∞=−21lim(ln(1)ln)xxxxx→+∞=+−i1limln(1)xxx→+∞=+1lim1xxx→+∞==2.2.2.2.求下列极限（1111）22201coslncoslimsinxxxxxeex−→−−−−;;;;解：原式222201122lim12xxxxx→+==−（2222）0ln()2sinlimsin(2tan2)sin(tan2)tanxxxexxxx→++−−;解：原式0ln(11)2sinlimsin(2tan2)sin(tan2)tanxxxexxxx→++−+=−−012sinlimsin(2tan2)sin(tan2)tanxxxexxxx→+−+=−−02lim442xxxxxxx→++==−−习题1-41-41-41-41111．求下列极限（1111）21lim(1sin)nnnn→∞−；4444/27272727解：原式2331111lim[1(())]3!nnnonnn→∞=−−+11lim((1))3!6no→∞=+=（2222）求33601limsinxxexx→−−；解：原式3636336600()112limlim2xxxxxoxxexxx→→++−−−===（3333）21lim[ln(1)]xxxx→∞−+；解：原式222111lim[(())]2xxxoxxx→∞=−−+12=（4444）21lim(1)xxxex−→+∞+；解：原式211[ln(1)]2limxxxxee+−−→∞==此题已换3333．设()fx在0x=处可导，(0)0f≠，(0)0f′≠.若()(2)(0)afhbfhf+−在0h→时是比h高阶的无穷小，试确定,ab的值.解：因为()(0)(0)()fhffhoh′=++，(2)(0)2(0)()fhffhoh′=++所以00()(2)2(0)(1)(0)(2)(0)()0limlimhhafhbfhfabfabfohhh→→′+−+−+++==从而10ab+−=20ab+=解得：2,1ab==−3333．设()fx在0x处二阶可导，用泰勒公式求00020()2()()limhfxhfxfxhh→+−+−解：原式222200001000220''()''()()'()()2()()'()()2!2!limhfxfxfxfxhhohfxfxfxhhohh→+++−+−++=22201220''()()()limhfxhohohh→++=0''()fx=4.4.4.4.设()fx在0x=处可导，且20sin()lim()2.xxfxxx→+=求(0),(0)ff′和01()limxfxx→+.解因为2200sin()sin()2lim()limxxxfxxxfxxxx→→+=+=[]220()(0)(0)()limxxoxxffxoxx→′++++=2220(1(0))(0)()limxfxfxoxx→′+++=5555/27272727所以1(0)0,(0)2ff′+==,即(0)1,(0)2ff′=−=所以01()limxfxx→+01(0)(0)()limxffxoxx→′+++=02()lim2xxoxx→+==习题1-51.计算下列极限(1)1112limnnn→∞+++⋯;;解：原式11lim1nnnn→∞+=+−1lim21nnnn→∞++==+(2)2212lim(1)nnnaanaana+→∞+++⋅⋅⋅+解：原式21lim(1)nnnnnanana++→∞=−−2lim(1)nnnana→∞=−−21aa=−2．设limnnaa→∞=，求(1)1222limnnaanan→∞+++⋯；解：原式22lim(1)nnnann→∞=−−lim212nnnaan→∞==−(2)12lim111nnnaaa→∞+++⋯，0,1,2,,.iain≠=⋯解：由于1211111limlimnnnnaaanaa→∞→∞+++==⋯，所以12lim111nnnaaaa→∞=+++⋯3．设2lim()0nnnxx−→∞−=，求limnnxn→∞和1limnnnxxn−→∞−.解：因为2lim()0nnnxx−→∞−=，所以222lim()0nnnxx−→∞−=且2121lim()0nnnxx+−→∞−=从而有stolz定理2222limlim022nnnnnxxxn−→∞→∞−==，且212121limlim0212nnnnnxxxn++−→∞→∞−==+所以lim0nnxn→∞=，111limlimlim01nnnnnnnxxxxnnnnn−−→∞→∞→∞−−=−=−6666/272727274．设110xq，其中01q≤，并且1(1)nnnxxqx+=−，证明：1limnnnxq→∞=.证明：因110xq，所以211211(1)111(1)()24qxqxxxqxqqq+−=−≤=，所以210xq，用数学归纳法易证，10nxq。又111nnnxqxx+=−，从而nx单调递减，由单调有界原理，limnnx→∞存在，记limnnxL→∞=在1(1)nnnxxqx+=−两边令n→∞，可得lim0nnx→∞=所以11limlimlim111nnnnnnnnnxxxx→∞→∞→∞+==−11(1)limlim(1)nnnnnnnnnnnnxxxxqxxxxxqx+→∞→∞+−==−−−11limnnqxqq→∞−==习题1-61-61-61-61.1.1.1.设()fx在(,)a+∞内可导，且()lim,xfxAx→+∞=lim()xfx→+∞′存在.证明:lim().xfxA→+∞′=证明：()()limlimlim()1xxxfxfxAfxx→+∞→+∞→+∞′′===2.2.2.2.设()fx在(,)a+∞上可微，lim()xfx→+∞和lim()xfx→+∞′存在.证明:lim()0xfx→+∞′=.证明：记lim()xfxA→+∞=（有限），lim()xfxB→+∞′=（有限），则()()()lim()limlimxxxxxxxxefxefxefxAfxABee→+∞→+∞→+∞′+====+从而0B=所以lim()0xfx→+∞′=3.3.3.3.设()fx在(,)a+∞上可导，对任意的0α,lim[()()]xfxxfxαβ→+∞′+=，证明:lim()xfxβα→+∞=.7777/27272727证明：因为0α，所以limxxα→+∞=+∞，由广义罗必达法则得()lim()limxxxfxfxxαα→+∞→+∞=11()()limxxfxxfxxααααα−−→+∞′+=lim[()()]xxfxfxα→+∞′=+βα=4444．设()fx在(,)a+∞上存在有界的导函数，证明:()lim0lnxfxxx→+∞=.证明：()()limlimlnln1xxfxfxxxx→+∞→+∞′=+，()fx′有界，1lim0ln1xx→+∞=+，所以()()limlim0lnln1xxfxfxxxx→+∞→+∞′==+习题2-1（此题已换）1.若自然数n不是完全平方数，证明n是无理数.1.证明3是无理数证明：反证法.假若3qp=,,(Nqp∈且qp,互质)，于是由223pq=可知,2q是2p的因子，从而得12=q即23p=,这与假设矛盾2.求下列数集的上、下确界.（1）11,nNn⎧⎫−∈⎨⎬⎩⎭解：1,inf0supEE==（2）