32

习题321用洛必达法则求下列极限(1)xxx)1ln(lim0(2)xeexxxsinlim0(3)axaxaxsinsinlim(4)xxx5tan3sinlim(5)22)2(sinlnlimxxx(6)nnmmaxaxaxlim(7)xxx2tanln7tanlnlim0(8)xxx3tantanlim2(9)xarcxxcot)11ln(lim(10)xxxxcossec)1ln(lim20(11)xxx2cotlim0(12)2120limxxex(13))1112(lim21xxx(14)xxxa)1(lim(15)xxxsin0lim(16)xxxtan0)1(lim解(1)111lim111lim)1ln(lim000xxxxxxx(2)2coslimsinlim00xeexeexxxxxx(3)axaxaxaxaxcos1coslimsinsinlim(4)535sec53cos3lim5tan3sinlim2xxxxxx(5)812csclim41)2()2(2cotlim)2(sinlnlim22222xxxxxxxx(6)nmnmnmaxnnmmaxanmnamxnxmxaxax1111limlim(7)22sec2tan177sec7tan1lim2tanln7tanlnlim2200xxxxxxxx177sec22seclim277tan2tanlim272200xxxxxx(8)xxxxxxxxx2222222cos3coslim3133secseclim3tantanlim)sin(cos23)3sin(3cos2lim312xxxxxxxxcos3coslim23sin3sin3lim2xxx(9)22221lim11)1(111limcotarc)11ln(limxxxxxxxxxxx122lim212limxxxx(10)xxxxxxxxxxx22022020cos1limcos1)1ln(coslimcossec)1ln(lim1sinlim)sin(cos22lim00xxxxxxx(注cosxln(1x2)~x2)(11)2122sec1lim2tanlim2cotlim2000xxxxxxxx(12)1limlim1limlim21012022ttttxxxxetexeex(注当x0时21xt(13)2121lim11lim1112lim12121xxxxxxxx(14)因为)1ln(lim)1(limxaxxxxexa而221)(11lim1)1ln(lim)1(ln(limxxaxaxxaxaxxxxaaaxaxxx1limlim所以axaxxxxeexa)1ln(lim)1(lim(15)因为xxxxxexlnsin0sin0limlim而xxxxxxxxxxcotcsc1limcsclnlimlnsinlim0000cossinlim20xxxx所以1limlim0lnsin0sin0eexxxxxx(16)因为xxxxexlntantan0)1(lim而xxxxxxxxx2000csc1limcotlnlimlntanlim0sinlim20xxx所以1lim)1(lim0lntan0tan0eexxxxxx2验证极限xxxxsinlim存在但不能用洛必达法则得出解1)sin1(limsinlimxxxxxxx极限xxxxsinlim是存在的但)cos1(lim1cos1lim)()sin(limxxxxxxxx不存在不能用洛必达法则3验证极限xxxxsin1sinlim20存在但不能用洛必达法则得出解0011sinsinlimsin1sinlim020xxxxxxxxx极限xxxxsin1sinlim20是存在的但xxxxxxxxxcos1cos1sin2lim)(sin)1sin(lim020不存在不能用洛必达法则4讨论函数00])1([)(2111xexexxfxx在点x0处的连续性解21)0(ef)0(lim)(lim212100feexfxx因为]1)1ln(1[101100lim])1([lim)(limxxxxxxxxeexxf而200)1ln(lim]1)1ln(1[1limxxxxxxxx21)1(21lim2111lim00xxxxx所以]1)1ln(1[101100lim])1([lim)(limxxxxxxxxeexxf)0(21fe因此f(x)在点x0处连续