洛必达法则例题

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)xxx5tan3sinlimπ→;(5)22)2(sinlnlimxxx−→ππ;(6)nnmmaxaxax−−→lim;(7)xxx2tanln7tanlnlim0+→;(8)xxx3tantanlim2π→;(9)xarcxxcot)11ln(lim++∞→;(10)xxxxcossec)1ln(lim20−+→;(11)xxx2cotlim0→;(12)2120limxxex→;(13)−−−→1112lim21xxx;(14)xxxa)1(lim+∞→;(15)xxxsin0lim+→;(16)xxxtan0)1(lim+→.解(1)111lim111lim)1ln(lim000=+=+=+→→→xxxxxxx.(2)2coslimsinlim00=+=−−→−→xeexeexxxxxx.(3)axaxaxaxaxcos1coslimsinsinlim==−−→→.(4)535sec53cos3lim5tan3sinlim2−==→→xxxxxxππ.(5)812csclim41)2()2(2cotlim)2(sinlnlim22222−=−−−=−⋅−=−→→→xxxxxxxxπππππ.(6)nmnmnmaxnnmmaxanmnamxnxmxaxax−−−−−→→===−−1111limlim.(7)177sec22seclim277tan2tanlim2722sec2tan177sec7tan1lim2tanln7tanlnlim22002200=⋅⋅==⋅⋅⋅⋅=+→+→+→+→xxxxxxxxxxxxxx.(8))sin(cos23)3sin(3cos2lim31cos3coslim3133secseclim3tantanlim22222222xxxxxxxxxxxxxx−⋅−==⋅=→→→→ππππ3sin3sin3limcos3coslim22=−−−=−=→→xxxxxxππ.(9)122lim212lim1lim11)1(111limcotarc)11ln(lim2222==+=++=+−⋅+=++∞→+∞→+∞→+∞→+∞→xxxxxxxxxxxxxxx.(10)xxxxxxxxxxx22022020cos1limcos1)1ln(coslimcossec)1ln(lim−=−+=−+→→→(注:cosx⋅ln(1+x2)~x2)1sinlim)sin(cos22lim00==−−=→→xxxxxxx.(11)2122sec1lim2tanlim2cotlim2000=⋅==→→→xxxxxxxx.(12)+∞====+∞→+∞→→→1limlim1limlim21012022ttttxxxxetexeex(注:当x→0时,+∞→=21xt).(13)2121lim11lim1112lim12121−=−=−−=−−−→→→xxxxxxxx.(14)因为)1ln(lim)1(limxaxxxxexa+∞→∞→=+,而aaaxaxxxaxaxxaxaxxxxxx==+=−−⋅+=+=+∞→∞→∞→∞→∞→1limlim1)(11lim1)1ln(lim)1(ln(lim22,所以axaxxxxeexa==++∞→∞→)1ln(lim)1(lim..(15)因为xxxxxexlnsin0sin0limlim+→+→=,而0cossinlimcotcsc1limcsclnlimlnsinlim20000=−=⋅−==+→+→+→+→xxxxxxxxxxxxxx,所以1limlim0lnsin0sin0===+→+→eexxxxxx.(16)因为xxxxexlntantan0)1(lim−+→=,而0sinlimcsc1limcotlnlimlntanlim202000=−=−==+→+→+→+→xxxxxxxxxxxx,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21)1(21lim2111lim)1ln(lim]1)1ln(1[1lim00200−=+−=−+=−+=−++→+→+→+→xxxxxxxxxxxxx,所以)0(lim])1([lim)(lim21]1)1ln(1[101100feeexxfxxxxxxxx===+=−−+−→−→+→.因此f(x)在点x=0处连续.