第一章：函数极限连续.pdf

高等数学辅导讲义习题讲解武 忠 祥)(xf1,0,1 x【解解】在在处处没没定定义义，xxxexxxxxfxxxxxxln)1(1limln)1(1lim)(limln111 11limln)1(lnlim11xxxxxxxxxxxexxxxxfxxxxxxln)1(1limln)1(1lim)(limln000 111limln)1(lnlim00 xxxxxxxxxxxexxxxxfxxxxxxln)1(1limln)1(1lim)(limln111 2111limln)1(lnlim11 xxxxxxxx ,0 b)(xf.0 x【解解】应应选选（C C）否否则则只只有有一一个个间间断断点点由由题题设设可可知知，0 x)(xf.1 x.eb 显显然然是是的的一一个个间间断断点点,而而另另一一个个间间断断点点只只能能是是而而,)(lim20eaxfx .0)(lim0 xfxeexaxxfxxx 12211)1)(lim)(limeexaxx 112)1(lim)1(eaexaxx21212111lim)1(【解解】0,01,1,2,1,0)(xxxexfx 故故应应选选（D D）.,0,0,10,0lim xxxenxn.11,11,1,0lim xxxxxnn不不存存在在，112lim)()1(nnxxnnxeexfxnxnxnxxneeexee21)(112lim 当当0 x时时)1ln()21ln()11ln(ln222nnnnxn 0 x,)1ln(1xxxx 【解解】当当时时，则则222222)1ln(1nknknknkknknnk nknnknkxnnk1212ln21)1(21limlim212 nnnnknnkn21)1(21limlim212 nnnnnnknnkn21lnlim nnx.lim21exnn 则则 )21(21limnnnn2)1(21lim2 nnnn【解解】)21(11lim2nnnnnnnn 322210 dxx【解解1 1】nxxxnxxxeexex2)2sin1ln(02120222lim)2sin1(lim nxxxxxee1lim2222)2sin1ln(02 222022)2sin1ln(lim nxxxxe12202)2(42sin12cos4lim nxxnxxxxe 02sin12coslim242202 axxxnenx22,2ean 222022)2sin1ln(lim nxxxxe2222222022)2(sin22sin2sinlim nxxxxxxe 0 a242202222sinlim,2exxeanx 【解解2 2】左左端端 222022)2sin1ln(lim nxxxxe【解解3 3】左左端端 22222022sin22sin)2sin1ln(lim nxxxxxxe2322202)2(61)2(sin21lim nxxxxe)61sin,21)1ln(32xxxxxx 2440222lim,2exxeanx 【解解1 1】原原式式)sin1ln()1ln()sin1ln()1ln(lim22220 xxxxx 4220)sin1ln()1ln(limxxxx 42220sin1sin1lnlimxxxxx )sin1(sinlim24220 xxxxx 300sinlimsinlimxxxxxxxx 3161lim2330 xxx)sin1ln()1ln()sin1ln()1ln(lim22220 xxxxx 4220)sin(11limxxxx 31)sin)(sin(lim40 xxxxxx【解解2 2】原原式式 (拉拉格格朗朗日日中中值值定定理理)4220)sin1ln()1ln(limxxxx 301)sin(limxxxxxxx 3sinln01elimxxxxx 【解解1 1】原原式式 )1lim(0 xxx20sinlnlimxxxx 30sinlimxxxx .61 20)sinln(1limxxxxx 33061limxxx 【解解2 2】原原式式xxxxxxxexln)1ln(ln111lim)1(lim xexxxeln)1ln(lnlim ,ln11eelimln)1ln(elim2lnlnlnxxxxxxxxxxxx 【解解】因因为为 x0lnxx.1lnln1limln1lnelimln)1ln(elimlnln xxxxxxxxxxxxxx.e)1(lim1ln11 xxxx而而当当时时，故故所所以以 【解解】)(11)2(11)1(111222nnnnn 122111222222nnnnnnnn )(11)2(11)1(1112222nnnnnn )(11)2(11)1(111lim222nnnnnn dxx 102114|ln1lim,11xxxxx1 x【解解】则则为为无无穷穷间间断断点点.111lim|ln)1(lim,1)1()1(xxxxxx111lim|ln1lim)1()1(xxxxx1 x为为跳跳跃跃间间断断点点.0|ln1lim,00 xxxxx0 x 为为可可去去间间断断点点.