函数与数列的极限的强化练习题答案(2)

2020-04-15 13:42

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sinxx?sinx?1 B. lim不存在

x??x??x?sinxx1?C. limxsin?1 D. limarctanx?

x??x??x2A. lim4.若limx?0f?2x??2, x1?tx1sintlim解:?limxsin ?选C

x??t?0xt则limx?0x? ( )

f?3x?11 C.2 D. 32A.3 B.

sinxsinxx?1?0?1 ?0;Blim注:Alimx??xx??sinx1?01?x1?2. 下列极限正确的是( )

2tx3x?2tlim3 解:limx?0f?3x?t?0f?2t??21211lim??? 3t?0f?2t?323te?0 B. lim?e?0 A. lim?x?0x?01x1xC. lim(1?cosx)x?0secx?e

?选B

?1?xsinx(x?0)??0(x?0)5.设f?x???且limf?x?存

x?01?xsin?a(x?0)?x??1xD. lim(?x??1x)?e

??e?e解:?lim?x?01x?1?0 ?选A ?e注:B:??,C:2,D:1

3. 若limf?x???,limg?x???,则下列在,则a= ( )

x?x0x?x0A.-1 B.0 C.1 D.2

sinx正确的是 ( )

?1, 解:?lim?x?0xA. lim??f?x??g?x?????

x?x0B. lim?f?x??g?x????? x?x0???1??limxsin??a??o?a ??x?0??x????a?1 选C

?6.当x?0时,f?x??1?x?1是比x高阶

a1?0C. lim

x?x0f?x??g?x?D. limkf?x????k?0?

x?x0?limkf?x??klimf?x??k??解:

x?x0x?x0k?0无穷小,则 ( )

A.a?1 B.a?0 C.a为任意实数 D.a?1

? ?选D

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a1解:lim1?x?1x?0?x?lim2xaa?1x?0?x0?a?1 故选A

二 、填空题(每小题4分,共24分)

x7.lim?x???x??1?x??? x解:原式

1?lim?x????1?1?x?1???exlim?x??1?x?e?1 8.lim?2?x?1?1?x?1?x2?1??? 解:原式

?????limx?1?2x?1?x?1??x?1? ?lim11x?1x?1?2

9.lim?2x?1?3?3x?2?97x???3x?1?100?

????解:原式

????397lim?x???2x?1??3x?1???lim?3x?2?x????3x?1?? ???2?3?3???827 .已知limx210?ax?6x?11?x存在,

则a= 解:?limx?1?1?x??0

?limx?1?x2?ax?6??0

1?a?6?0,a??7

11.lim?1x?0???exsin1arcsinx?x2?x??? 11解:?sin1xlimx?0?ex?0?limx?0exsin12?1,x2?0又?limarcsinxx?0x?limxx?0x?1 故 原式=1

12.若limx2ln?1?x2?x?0sinnx?0

且limsinnxx?01?cosx?0,则正整数n= 解:?limx2ln?1?x2?x?0?limx2?x2sinnxx?0xn n?40,limxnn?2x?0x20?n?2,n?4, 故2n?3

三、计算题(每小题8分,共64分)

13.求limsin3x?2xx??sin2x?3x

sin3x解: 原式=limx?2x??sin2x x?3?limsin3xx??x?0???sin3x?1,lim1?x??x?0??

limsin2xx??x?0???sin2x?1,lim1x??x?0???

?原式?0?20?3??23 14.求lim1?tanx?1?sinxx?0x?1?cosx?

7

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解:原式

有理化

limtanx?sinxx?0x(1?cosx)(1?tanx?1?sinx) ?limtanx(1?cosx)x?0x(1?cosx)?12 ?limtanx11x1x??x?2?2limx?0x?2

15.求lim?x????sin2xx?cos1?x??

解:令

1x?t,当x??时,t?0 ?1原式?limcost?sin2t?tt?0 1?limt?0?1?cost?1?sin2t?t

???cost?1?sin2telimt?0t?e2

16.求limlncos2xx?0lncos3x

解:原式

变形limln?1?cos2x?1?x?0ln?1?cos3x?1?

等价limcos2x?1x?0cos3x?1

?1等价22lim2?x?x?0?4 ?12?3x?29????注:原式

????lim?2sin2xcos3x?0cos2x?x?3sin3x

????49 17.求limex?e?x?2xx?0x?sinx

0 解: 原式

0limex?e?x?2x?01?cosx 000ex?e?xlim0ex?e?xx?0sinxlimx?0cosx?2 ??1f?x????ex?a,x?018.设?且lim?1?cosxx?0f?x?存在,

??x,x?0求a的值。

解:??xlim?0??e?1x?a???e???a?0?a?a

??x2lim1?cosxx?lim2x?0?x?0?x1?lim2??x?x?0?x??22 ?a??22 119.1?3lnxxlim?0??sin3x?

解: 原式

?00?换底法lim3cosxln(sin3x)x?0?sin3xelim3x?0?1?3lnx?ex

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?ex?0?3sinxlim3x?ex?0?3xlimx?e

13x3(3)x?sinx等价于?x?0?

620.求lim?x?x2ln?1?x??????1???? x??解: 原式

1?txx3(4)arcsinx?x等价于?x?0?

6证:?1??limx?0?1ln?1?t??lim??? 2t?0tt??

通分limt?0t?ln?1?t?t2tanx?sinx

x32?0????0??0????0?11?lim1?t t?02tlimx?0tanx?1?cosx?x32

?limt?01?t?111?lim?

2t?t?1?t?0t?12四、证明题(共18分) 21.当x??时且

x2x??lim32?1 x?0x2limu?x??0,limv?x???,

x??x??x3当x?0时,tanx?sinx?

2

证明lim??1?u?x???x??v?x??e

x??limu?x?v?x?证:lim??1?u?x???x??v?x?tanx?xsec2x?1?lim ?2??limx?0x?013x2x3?lim??1?u?x???x??1?u?x??v?x?u?x?

x2证毕 当x?0时,tanx?x?

322.当x?0时,证明以下四个差函数的等价无

穷小。

?ex??limu?x??v?x?tan2xx2?lim2?lim2?1 x?0x?0xx

x3(1)tanx?sinx等价于?x?0?

2x3(2)tanx?x等价于?x?0?

3

9

?3??limx?0x?sinx1?cosx ?limx?01312xx62专业精神 诚信教育 同方专转本高等数学内部教材 严禁翻印

12x?lim2?1 x?012x2当x?0时,x?sinx?2lim?x???2?n?x?2n???0 x?2?lim?x2?mx?8??0,4?2m?8?0

x?213x 62m?12,m?6

arcsinx?x ?4??limx?013x6x2?6x?8(2)?lim2

x?2x?2?nx?2n???0????0?121?x?limx?012x2?11?1?x2?lim x?0122x1?x2lim2x?64?6 ?x?22x??2?n?4??2?n???21? 2?n512x?lim2?1 x?012x?12当x?0时,arcsinx?x等价于??10?2?n n?12 答m?6,n?12

选做题

113x 61?x1?x??求

lim?x?0?e???x? ???1x1x五、综合题(每小题10分,共20分) 23.求lim3x?9x2?12x?1

x????解:原式1???1?x??e??? lim1?x?0??e????1?x???1?x??lim?x?0e解: 原式

有理化limx??9x2??9x2?2x?1?3x?9x?2x?1 2

?limx???2x?13x?9x?2x?12?ex?0lim?1?x?x?ex?e1??????e1x

令y??1?x??e1x1ln?1?x?x1?2??21x?lim???

x??3213?33?9??2xxy???1?x?11x?ln?1?x?1?x 2xx2?1?x?

??1?x?xx??1?x?ln?1?x?x??1?x?ln?1?x?x2?1?x?12x2?mx?8124. 已知lim2?,求常数

x?2x??2?n?x?2n5m,n的值。

解:(1)∵原极限存在且

原式?elimx?0lim?ex?0lim0?ln?1?x?2x?3x2

?x2?ex?02x?3x?e

10

?


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