又当x→0进，
1?cosax~1221ax,1?cosbx~b2x2,22所以
122axlncosaxcosax?11?cosaxa22lim?lim?lim?lim?2.x?0lncosbxx?0cosbx?1x?01?cosbxx?0122bbx2
22sinxx?0,?0x2xx?0ee(14)因为当时，
22?sin2x?sinx?x2?xln?1?x?~,ln?1?2x?~2x,xee??e?e 故 ?所以
?sin2x?ln?1?x?ln(sin2x?ex)?xln(sin2x?ex)?lnexe?lim?lim?lim?22x22x2xx?0ln(x?e)?2xx?0ln(x?e)?lnex?0?x2?ln?1?2x??e?sin2x22xsinxsinx????xx?lime2?lime??e??lim??lim?x?0x?0x?0x?x??x?0x?e2x?e0?1?1.
1??1????1???1?(1)lim?1???lim??1?????lim?1????e2?e.x???x??x???x???x???x??29. 解：
xxx21212?x?3?(2)lim??x???x?2?2x?15???lim?1??x???x?2?2x?1x?25??5?5?5???1??lim????1??x???x?2????x?2????5???5??10510?e?1?e.1????lim???x???x?2?????
13310?5???lim?1?????x???x?2?(3)lim(1?3tanx)x?03x?21052cot2x1?????lim(1?3tan2x)3tan2x??lim(1?3tan2x)3tan2x??e3.?x?0????x?0? 3x2lncos2x(4)lim(cos2x)x?0x2?limex?0?limex?01????ln?cos2x?1?21?(cos2x?1)??x????3cos2x?13(cos2x?1)?limex?0x2ln?1?(cos2x?1)?1cos2x?1?e?e?e3limcos2x?1x2?2sin2xx2x?0?limln?1?(cos2x?1)?x?01cos2x?13limx?01????cos2x?1??ln?lim?1?(cos2x?1)????x?0?2sinx??6???lim??lne?x?0x??e?6?1?1?e?6.2
精选
x2?x?2?(5)limx[ln(2?x)?lnx]?lim2??ln?lim2ln?1??x??x??x??2x?x?x2x2x??2??22???2limln?1???2?ln?lim?1????x???x???x??x????2lne?2.
(6)令x?1?t，则当x?1时，t?0.
1?xt111lim??lim????????1.11x?1lnxt?0ln(1?t)lne??ln?lim(1?t)t?limln(1?t)tt?0?t?0?
11lny?ln(ex?x)xxx30. 解：(1)令y?(e?x)，则
于是：
x?lnex?ln?1??x?11x?e??x?xlimlny?limln?e?x??limlne?1?x??limx?0x?0xx?0xx?e?x?0
1?ex1?x?x???lim?1??xln?1?x??1?lim?limln1?????x?0x?0exx?0xe?ex??e????1?1?lne?2
即
exxy?2lnlimx?0?? 即x?0xxxlimy?e1x2 即
lim?e?x??e2x?0x1x.
xxx?a?b?c?1a?b?cy??lny?ln?3??x3(2)令，则
于是
1ax?bx?cxlim(lny)?limlnx?0x?0x331??ax?bx?cx?3?ax?bx?cx?3??limln?1????x?0x??3????xxxxxax?bx?cx?33即x?0a?b?c?3?a?b?c?3?a?b?cx?3?lim?limln?1??x?0x?03x3??31?ax?1bx?1cx?1???ax?bx?cx?3?ax?bx?cx?3??lim??????ln?lim?1??3x?0?xx?0xx???3????1?(lna?lnb?lnc)?lne?ln3abc3
lim(lny)?ln3abc,limy?3abclnlimy?ln3abc,xxx3 即
?x?0? 故x?0 ?a?b?c?lim??3abc?x?03??即 .
xxx1x精选
11?11???lny?xln?sin?cos?y??sin?cos?xx? ?xx?，则?(3)令
于是
x??11????11?sin?cos?1limlny?limxln?1??sin?cos?1?xx??x??x????xx????????11??1?sin?cos?1?x??x即x??11???11????limx?sin?cos?1??ln?1??sin?cos?1??x???xx???xx??11??1sin1?cos????11??xx?11?sin?cos?1???lim????ln?lim?1??sin?cos?1??xx?x??11xx???????x?????x?x?2?1?1??1sin????2x?lim?x???lne?(1?0)?lne?1??limx??1??x??1??xx??
limlny?1limy?elnlimy?1 从而
111sin?cos?1xx
?x??? 故x??x
1??1lim?sin?cos??ex???xx?即 .
1?1???lny?xln?1?2?y??1?2??x? ?x?，则(4)令
于是：
1x?x2??1??1lim(lny)?limxln?1?2??limxln??1???2?x??x???x?x????x??2x即
1?11?1???limln?1?2??lim?limln?1?2?x??xx??xx???x??x??0?lne?0
y?0lim(lny)?0, lnlimx??x??x2x2??
1??lim?11???limy?12?x???x?x?? 即.
x?xxx(1)limf(x)?lim?lim?1,limf(x)?lim?lim??1??????x?0x?0x?0x?0x?0x?0xxxx31.解：
limf(x)?limf(x)??因为 所以x?0(2)x?2x?0x?0x
limf(x)不存在.
x?2limf(x)?lim??1???, limf(x)?lim(x?2)?4x?2?x?2?x?2
精选
因为
x?2?limf(x)不存在，所以x?2limf(x)不存在.
32. 解：(1)由初等函数的连续性知，f(x)在（0，1），（1，2）内连续， 又
2Qlimf(x)?lim(2?x)?1, limf(x)?limx?1????x?1x?1x?1x?1
?limf(x)?1,x?1 而f(1)?1，?f(x)在x?1处连续，
x?0又，由x?02limf(x)?limx?0?f(0)??，知f(x)在x?0处右连续，
综上所述，函数f(x)在[0，2)内连续. 函数图形如下：
图1-2
(2) 由初等函数的连续性知f(x)在(??,?1),(?1,1),(1,??)内连续，又由
x??1?limf(x)?lim?1?1, lim?f(x)?lim?x??1,x??1x??1x??1
知
x??1?limf(x)又由x?1?不存在，于是f(x)在x??1处不连续.
x?1x?1x?1limf(x)?limx?1, limf(x)?lim1?1,???
limf(x)?f(1)及f(1)?1知x?1，从而f(x)在x=1处连续，
综上所述，函数f(x)在(??,?1)及(?1,??)内连续，在x??1处间断.函数图形如下：
图1-3
nx?n?xn2x?1f(x)?limx?lim2x??1,n??n?n?xn??n?1 (3)∵当x<0时， n0?n0f(x)?lim0?0,n??n?n0当x=0时，
1n?nn?1n2x?1f(x)?limx?lim?limn??n?n?xn??n2x?1n??11?2xn当x>0时，
??1,x?0,x?xn?n??f(x)?limx??0,x?0,n??n?n?x?1,x?0. ?由初等函数的连续性知f(x)在(??,0),(0,??)内连续，
x?x2x1?精选
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