3sinx3ln(1?2x)sinx6xln(1?2x)2xsinx1(15)lim(1?2x)x?0?limex?0?limex?0?eu?xlim6?x?0sinx12?ln(1?2x)x?ex6?lim?limln(1?2x)2xx?0sinxx?0?e6?1?lne?e6.sinxsinxlimu?lim?1x?0x?0x， 则x(16)令
sinxlimln?0.limlnu?0x?0u?1x而 所以
x2?x3x?x2?lim?lim?02x?02?x26. 解：x?02x?x
232∴当x?0时，x?x是比2x?x高阶的无穷小量.
1?x11(1)?lim?lim?x?11?x2x?11?x2 27.解：
∴当x?1时，1?x是与1?x同阶的无穷小.
21(1?x2)1?x(2)?lim2?lim?1x?1x?11?x2
1(1?x2)∴当x?1时，1?x是与2等价的无穷小. 28. 解：(1)因为当x?0时，sinmx~mx,sinnx~nx,
sinmxmxm?lim?.n 所以x?0sinnxx?0nxlimcosxxcosxlimx?0(2)limxcotx?lim?cosx?lim??1.x?0x?0sinxx?0sinxsinxlimx?0xx1?cos2x2sin2xsinx(3)lim?lim?2lim?2.x?0x?0x?0xsinxxsinxx
(4)因为当x?0时，
ln(1?exsin2x)~exsin2x,1?x2?1~12x2,所以
2exsin2x?sinx?lim?lim?lim2ex?lim???2.2x?0x?0x?0x?0?12x?1?x?1x2
(5)因为当x?0时，arctan3x~3x,所以
ln(1?exsin2x)arctan3x3x?lim?3x?0x?0xx. xxsinnsinnx2?xlim2?x.(6)lim2nsinn?limx?n??n??n??xx22n2n
1x?2时，arcsin(1?2x)~1?2x,所以 (7)因为当
lim
4x2?14x2?1(2x?1)(2x?1)lim?lim?lim??lim(2x?1)??2.1arcsin(1?2x)11?2x111?2xx?x?x?x?2222
(8)因为当x?0时，
arctanx2~x2,sinxx~,arcsinx~x,22所以
arctanx2x2lim?lim?2x?0x?0xxsinarcsinx?x22.
1sinx~x,1?cosx~x2,sinx3~x32(9)因为当x?0时，,所以
1x?x2tanx?sinxsinx(1?cosx)lim?lim?lim3233x?0x?0x?0x?cosxsinxsinxcosx11?lim?.x?02cosx2 ????????????sinx~x,sinx~x2222(10)因为当x?0时，，所以
???????2sinxsinxcos?x?cos?x22lim?lim22x?0x?0xx???????2?x?x22?lim2x?0x1?(?2??2).2 xxarcsin~,ln(1?x)~?x,221?x1?x(11)因为当x?0时，所以 xxarcsin2211?x1?xlim?lim??lim??1.2x?0x?0x?0ln(1?x)?x1?x
(12)因为当x?0时，sinx~x,sin2x~2x,所以
1?cos4x2sin22xlim?lim2x?02sin2x?xtan2xx?0sinx(2?xsec2x)2?(2x)28?lim2?limx?0x(2?xsec2x)x?02?xsec2x8??4.2lim(2?xsecx)x?0(13)因为lncosax?ln[1?(cosax?1)],lncosbx?ln[1?(cosbx?1)], 而当x?0时，cosax?1?0,cosbx?1?0
1故 ln[?(caoxs?1)]~axc?os?1,ln[b1x?(cosb1)x? ]
又当x→0进，
1?cosax~1221ax,1?cosbx~b2x2,22所以
122ax2lncosaxcosax?11?cosaxalim?lim?lim?lim2?2.x?0lncosbxx?0cosbx?1x?01?cosbxx?0122bbx2
22sinxx?0,?0x2xe(14)因为当x?0时，e
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??27. 解：
xxx21212?x?3?(2)lim??x???x?2?2x?15???lim?1??x???x?2?2x?1x?25???5?55???1??lim????1???x??x?2?????x?2???10x?25???5???105105?5??lim1???lim??e?1?e.??1?????x??x?2????x???x?2?????
10(3)lim(1?3tan2x)cotx?02x1????223tan?lim(1?3tan2x)3tanx??lim(1?3tanx)2x??e3.?x?0????x?0?
13333(4)lim(cos2x)x?0x2?limex?0x2lncos2x?limex?01????lncos2x?1??2x??1?(cos2x?1)????3cos2x?13(cos2x?1)?limex?0x2ln?1?(cos2x?1)?1cos2x?11?e?e?ecos2x?1cos2x?13lim?limln?1?(cos2x?1)?2x?0x?0x3lim?2sin2xx2x?01????cos2x?1??ln?lim?1?(cos2x?1)??x?0???2sinx??6???lim??lne?x?0x??e?6?1?1?e?6.2
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?tln?limln(1?t)tlim(1?t)??t?0?t?0?
11lny?ln(ex?x)xxx28. 解：(1)令y?(e?x)，则
于是：
x?lnex?ln?1?x??11x???e?limlny?limln?ex?x??limlnex?1?x??limx?0x?0xx?0xx?e?x?0
1?ex1?x?x???lim?1??xln?1?x??1?lim?limln1????x?x?0x?0exx?0xee??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?0x33??xxxxx1a?b?cx?3??a?b?c?3?limln?1????x?0x??3????xxxxxax?bx?cx?33a?b?c?3?a?b?c?3?a?lim?limln?1??x?0x?03x3??x3x?b?cx?3x31?ax?1bx?1cx?1???ax?bx?cx?3?ax?bx?cx?3??lim??????ln?lim?1??3x?0?xxx???3??x?0??即x?01?(lna?lnb?lnc)?lne?ln3abc3lim(lny)?ln3abc,lnlimy?ln3abc, 即
?x?0?
故x?0limy?3abc
?a?b?c?lim??3abc?3?即 x?0?.
xxx1x
11?11???y??sin?cos?lny?xln?sin?cos?xx?，则xx? ??(3)令
于是
x??11????11?sin?cos?1?limlny?limxln?1??sin?cos?1?xx?x??x???xx?????????11??1?sin?cos?1?x??x111???11??sin1?cos1?1??limx?sin?cos?1??ln?1??sin?cos?1??xxx???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?lim?x???lne?(1?0)?lne?1x??1??x??1??xx??
y?1limlny?1limy?elnlimx?? 从而
即x???? 故x??x
11??lim?sin?cos??ex???xx?即 .
1?1???y??1?2?lny?xln?1?2??x?，则?x? (4)令
于是：
1x?x2??1??lim(lny)?limxln?1?2??limxln??1?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
lim(lny)?0, lnlimy?0x??x2x2?x???
1??lim?11???limy?12?x???x?x?? 即.
xx(1)lim?f(x)?lim??lim??1,limfx(?)?x?0x?0x?0x?0xx31.解：
limf(x)?limf(x)??因为 所以x?0(2)x?2x?0x?0xxlim??x?0x?x?xlim???0x
1
limf(x)不存在.
x?2limf(x)?lim??1???, limf(x)?lim(x?2)?4??x?2x?2x?2