11??xx④ limx??1?x??x?
x????21?1?1?ln1?x??lnx2?1解：原式?limx?ex?ex??limx?ln?1?x??lnx?
x???x???x?x???21??ln?1??x?? ?limx?ln1?x?lnx?lim?1 ?????x???x???1xx?1??2?cosx?⑤ lim3????1? x?0x3??????ln?2?cosx??ln31?2?cosx?解：原式?lim3ln? ?lim???2x?0xx?03x6??1x⑥ limcosx?ecosx?3?x2
cosxx?0cosx?1?e解：原式?lim12??x2?1x?0cosx?3cosx?，而1?cosx?1x22，e?x2?1，cosx?1，
?x?0?且limx?0122。故原式????1?21?x2x1x?xx222??6
123计算下列极限 ① lim1?cosx?lncosxex2x?0?e?x2?sinx22
解：1?cosx?故原式?1。
312x，e?ex2?x2?x???x2?2???2x2,sinx?x，且lim222x?x22x?0??2??1，
② limcosx?tgx?3cosx?sinxx?0ln?1?x??ln?1?sinx??sinx1 2? 解：原式=lim3x?0x??sinx?x3???tgx?sinx?③ limln?x?ex??2sinxx?0sin?2tgx??sin?tgx??tgx
解：原式可如下考虑：若x?0时，sin?2tg2x??sin?tg2x??tg2x?2x， 又?tgx??x,limx?e?12xx2x?xx?0??2??1，lnx?ex?x?ex?1?2sinx?2x
??且limx?0?1??1，故原式?limaxbxx?e?1?2x2x?xxx?0?4。
④ lim?1?ax???1?bx?x?0?tg2x?x??sinx?x?eax,a?b
bx解：原式?limln?1?ax??eln?1?bx?x?0?2x?x??x?x?
?a2ln?1?ax?b2ln?1?bx??a2?b2? ?lim? ??x?0ax2bx2?2?习题1—4
1??1求?lim?1??x???x??x2e?x
解：原式?limex???3?1?2xln?1???xx???limex???1?1??2?1x???0??x22???x???x2x?limex???12?1???x0?2?2?x??e?12
2 求limex?1?xsin6x3
x6x?01?x?3解：原式?lim2?0?xx66??1?x3x?0?12
?sinxf?x??03 设f?x?在x?0处可导，且lim?2?求f?0?,f????2，x?0xx??x?x3?及limx?01?f?x?x
?sinx?xf?x??解：因lim???lim2x?0x?0x??3!?o?x4??x?f?0??f??0?x?o?x??x2
x3?1?f?0?x?? ?lim故f?0???1,f??0??2
x?0f??0?x?22x3!?2
lim1?f?x?xx?0?lim1?f?0??f??0?x?o?x?xxx?0o?x?????lim?f?0????2 x?0x??4求limx?02
?xln?1?x??xex?1?1?解：原式?limx?02?xln?1?x??xe22x?1?1?x1?1???1?1x2?22?21?x?x?o?x??1?222 ?lim2x?0x2x??o?x??x?1???x??o?x??2??limx?0182x2x?o?x???14
?xo?x?n5 求lim?nn?1?n??lnn
?1????1?x??1?ln???x?xxlnxln?1?x?lime?e?x?0x?0?x?x1limx?01xn解：原式limn?1lnn1xn??1n?limx?0，而lim??e?1
0?1??1?故ln??????1?x?0?，故原式=1。
?x??x?xx6 设f?x?在x?0处可导，f?0??0,f??0??0,若af?h??bf?2h??f?0?在h?0时是比h高阶的无穷小，试确定a,b的值。
af?h??bf?2h??f?0?h解：limh?0?0
即 lima?f?0??f??0?h?o?h???b?f?0??f??0?2h?o?2h???f?0?hh?0?0
即 lim?a?b?1?f?0??f??0?h?a?2b??ao?h??bo?2h?hh?0?0
?a?b?1?0?a?2?故? ?a?2b?0b??1??7 求limn2?1?nsinn????1?? n?1?nsin1n?limx?sinx?1，利用sinx?x?x?ox4
??3x?03!x63解：原式?limn??1n22n?2?1??8求limn?e??1???
n??n??????1??2e??1??n??解：原式?limn??1n22n?lime??1?x?22x2x?0x?lime?e2xln?1?x?x?0x
?lime?exx2?x?ox?2?x?o?x??x?0?lime?ex2x2??limex?02x?ln?1?x?xx?0???2??e
2?n???1?9求limn?e?1???1?
n??n??????1x解：原式?lime?1?x?x??1?lime?e?1xln?1?x??1x?0x?0x?lime1?1xln?1?x??1x?0x
1??1?ln?1?x???1?1x?????0?e ?lime??x?0x210设f????x0?存在，求极限lim1n3n?0??f?x0?3h??3f?x0?2h??3f?x0?3h??f?x0???
12f???x0??3h??2解：f?x0?3h??f?x0??f??x0?3h?16f????x0??3h??o3??3h??
3类似可得f?x0?2h?,f?x0?h?的表达式代入化简，可将原式?f????x0??6。
习题1—5
1 列极限
1?12?????n1n① lim n??解：令xn?1?12?????1n,yn?n，利用stolz公式可得原式?2
② lim1?2?????nnn
n??解：令xn?1?22?????nn,yn?nn，利用stolz公式可得原式?0 ,a?1
1a?a2③ lim1?a?2a????nanan?2n??解：令xn?1?a?2a2?????nan,yn?nan?2，利用stolz公式可得原式?2设liman?a，求
n??
① lima1?2a2?????nann2
n??解：令yn?n2,xn?a1?2a2?????nan，则lim?n?1?an?12n???n?1??n2?12a，故原式?12a
② lim1nnn???k?1akk 解：令xn?a11a1?a22a22????annann，yn?n 原式?lim1?????nan?1?limn?1n?1?n?liman?1n??n??n???n?1n?1?n?2a
?③ lim1lnnnn???k?1akk
a11a22ann解：令yn?lnn,xn?a1a22??????
an?1an?1an?1?
1?e??n?1?ln?1??n??原式?lim1n????????annlimn??lnnn?1?limln?n?1??lnnn??④ limn1a1?1a2?????1ann??,ai?0,i?1,2,???n
1解：原式?lima1?1a2?????n1an?1an??，故原式?a