(1?x)(1?3x)?(1?nx)(10)limx?1(1?x)n?1 ?limx?1(1?x)n?1(1?x)n?1(1?x)(1?3x?3x2)(1?4x?4x2?4x3)?(1?nx?nx2???nxn?1)1 ?limx?1(1?x)(1?3x?3x2)(1?4x?4x2?4x3)?(1?nx?nx2???nxn?1) ?112?3?4???n?n!.(11)lim?1?x?x2?3x2?x?1?1?1?x?3?1?x3???limx?1(1?x)(1?x?x2)?limx?2x?1(1?x)(1?x?x2) ?lim(x?1)(x?2)
x?1(1?x)(1?x?x2)?lim?(x?2)x?11?x?x2??1.(x?1)2(12)?limlim(?1x?1)2x?1x2?x?1?xlim(2?0x?1x?x?1)
?limx2?x?1x?1(x?1)2??.(13)?log)1a(1?x?loga(1?x)xx
1而lim(1?x)x?e. 而limlog1x?0u?eau?logae?lna ?limloga(1?x)x?0x?1lna.
(14)令u?ax?1,则x?loga(1?u),当x?0时，u?0.
所以limax?1x?0x?limuu?0log?1?lna(利用(13)题的结果). a(1?u)limloga(1?u)u?0u336x(15)lim(1?2sinxx?0x)?limesinxln(1?2x)sinxx?0?lime2xln(1?2x)x?011?elimxx?06?sinx?ln(1?2x)2x?6?limxe?limln(1x?0?2x)2xx?0sinx
?e6?1?lne?e6.(16)令u?sinxx， 则limsinxx?0u?limx?0x?1 16
而limlnu?0 所以limlnu?1x?0sinx?0. x30. 当x?0时，2x?x2与x2?x3相比，哪个是高阶无穷小量？
x2?x3x?x2解：?limx?02x?x2?limx?02?x?0 ∴当x?0时，x2?x3是比2x?x2高阶的无穷小量.
31. 当x?1时，无穷小量1?x与(1)1?x2,(2)12(1?x2)是否同阶？是否等价？解：(1)?lim1?xx?11?x2?lim1x?11?x?12
∴当x?1时，1?x是与1?x2同阶的无穷小.
1(1?x2)(2)?lim2x?11?x?lim1?xx?12?1
∴当x?1时，1?x是与12(1?x2)等价的无穷小.
32. 利用limsinxx?0x?1或等价无穷小量求下列极限： (1)limsinmxx?0sinnx; (2)limx?0xcotx;1?cos2xln(1?ex(3)limsin2x)x?0xsinx; (4)limx?01?x2?1 (5)limarctan3xx?0x; (6)limn??2nsinx2n;4x2?1arctanx2(7)limx?1arcsin(1?2x); (8)limx?0;2sinx2arcsinx(9)limtanx?sinxx?0sinx3; (10)limcos?x?cos?xx?0x2;
arcsinx(11)lim1?x2x?0ln(1?x); (12)lim1?cos4xx?02sin2x?xtan2x;(13)limlncosaxln(sin2x?x?0lncosbx; (14)limex)?xx?0ln(x2?e2x)?2x.解：(1)因为当x?0时，sinmx~mx,sinnx~nx,
所以limsinmxx?0sinnx?limmxmx?0nx?n.
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cosxxcosxlim(2)limx?0xcotx?limx?0sinx?cosx?limx?0sinx?x?0?1.xlimsinxx?0x 1?cos2x2sin2(3)limx?0xsinx?limxsinxx?0xsinx?2limx?0x?2.(4)因为当x?0时，ln(1?exsin2x)~exsin2x,1?x2?1~12x2,所以 ln(1?exsin2x)2limx?01?limexsin2x?sinx??x2?1x?01?lim2ex?lim2x?0x?0??x???2. 2x(5)因为当x?0时，arctan3x~3x,所以
limarctan3xx?0x?lim3xx?0x?3.
(6)limxsinxx2nsinn??2nsin2n?limn??x?x?xlim2nn??x?x. 2n2n(7)因为当x?12时，arcsin(1?2x)~1?2x,所以
lim4x2?14x2?1(2x?1)(2x?1)x?12arcsin(1?2x)?limx?121?2x?limx?121?2x??lim(2x?1x?1)??2.2(8)因为当x?0时，arctanx2~x2,sinx2~x2,arcsinx~x,所以 limarctanx2x?0?limx2sinx2arcsinxx?0x?2. 2?x(9)因为当x?0时，sinx~x,1?cosx~12x2,sinx3~x3,所以 limtanx?sinxsinx(1?x?1x2x?0sinx?limcosx)x?0sinx3cosx?lim23x?0x3?cosx ?lim1x?02cosx?12.(10)因为当x?0时，sin?????2x~???2x,sin?2x~???2x，所以
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?2sin???xsin???limcos?x?cos?x22xx?0x2?limx?0x2?2????x?????lim22xx?0x2
?12(?2??2).(11)因为当x?0时，arcsinxx1?x2~1?x2,ln(1?x)~?x,所以
arcsinxxlim1?x2x?0ln(1?x)?lim1?x2?x??lim1x?0x?01?x2??1.
(12)因为当x?0时，sinx~x,sin2x~2x,所以
lim1?cos4xx?02sin2x?xtan2x?lim2sin22xx?0sin2x(2?xsec2x)?lim2?(2x)28x?0x2(2?xsec2x)?limx?02?xsec2x ?8lim(2?4.x?0?xsec2x)(13)因为lncosax?ln[1?(cosax?1)],lncosbx?ln[1?(cosbx?1)], 而当x?0时，cosax?1?0,cosbx?1?0
故 ln[1?(caoxs?1)]~axc?os?1,ln[1bx?(cosb1)x? ]又当x→0进，1?cosax~12212ax,1?cosbx~2b2x2,所以 1limlncosaxcosax?11?cosa2x2x?0lncosbx?limx?0cosbx?1?limaxx?01?cosbx?lim2x?01?a222b2.2bx0时，sin2(14)因为当x?xx2ex?0,e2x?0 故 ln??sin2x?sin2x?x2?x2?1?ex??~ex,ln??1?e2x??~e2x, 所以
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?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?????lime2?limex???limex??lim?? x?0x?0x?0x?0xx??x??e2x?e0?1?1.33. 利用重要极限lim(1?u)?e，求下列极限：
u?01u?1??x?3?(1)lim?1??; (2)lim??x???x???x?2?x?(3)lim(1?3tanx)x?0x??2cot2xx?0x22x?1;32; (4)lim(cos2x)x;
1?x.x?1lnxx12(5)limx[ln(2?x)?lnx]; (6)limx2x121??1????1??1??解：(1)lim?1???lim??1?????lim?1????e2?e.
x???x??x???x???x???x???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??1010x?25???5??105105?5??lim?1???lim??e?1?e. ??1?????x????x???x?2?????x?2??2(3)lim(1?3tan2x)cotx?0x1????223tan?lim(1?3tan2x)3tanx??lim(1?3tanx)2x??e3. ?x?0????x?0?1cos2x?13333(4)lim(cos2x)x?0x2?limex?0x2lncos2x?limex?01????lncos2x?1??21?(cos2x?1)??x????33(cos2x?1)?limex?0x2ln?1?(cos2x?1)?1cos2x?1?e?e?e3limcos2x?1x2?2sin2xx2x?0?limln?1?(cos2x?1)?x?01cos2x?1
3limx?01????cos2x?1??ln?lim?1?(cos2x?1)??x?0???2sinx??6???lim??lnex?0x???e?6?1?1?e?6.2 20