函数与数列的极限的强化练习题答案(2)

2018-11-20 19:08

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f?x?y??f?x??f?y?,证明f?y?为奇

函数。

解 (1)求f?0?:令

注:B:??,C:2,D:1

3. 若limf?x???,limg?x???,则

x?x0x?x0下列正确的是 ( ) A. lim??f?x??g?x?????

x?x0x?0,y?0,f?0??2f?0??f?0??0

(2)令x??y:f?0??f??y??f?y??f??y???f?y?

B. lim?f?x??g?x????? x?x0?C. limx?x0?f?y?为奇函数

1 ?0f?x??g?x?第二讲:函数的极限与洛必达法则的强化练习题答案

一、单项选择题(每小题4分,共24分) 1. 下列极限正确的( ) A. lim在

D. limkf?x????k?0?

x?x0解:?limkf?x??klimf?x??k??x?x0x?x0k?0?

sinxx?sinx?1 B. lim不存

x??x??xx?sinx?选D

4.若limx?01?C. limxsin?1 D. limarctanx?

x??x??x2f?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?2t3lim解:lim x?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

1xD. lim(?x??1x)?e

??e?e解:?lim?x?01x?1?0 ?选A e?6

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??1xsinx(x?0)?5.设f?x????0(x?0)且lim??xsin1?a(x?0)x?0f?x??x?存在,则a= ( ) A.-1 B.0 C.1 D.2 解:?limsinxx?0?x?1, xlim???0?????xsin1?x????a???o?a ?a?1 选C

6.当x?0?时,f?x??1?xa?1是比x高阶无穷小,则 ( )

A.a?1 B.a?0 C.a为任意实数 D.a?1a1xa解:lim1?x?12a?1x?0?x?limx?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

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

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

39.lim?2x?1??3x?2?97x???3x?1?100? ????解:原式

????397lim?x???2x?1??3x?1???lim?3x?2?x????3x?1?? ??3?2?8?3???27 10.已知limx2?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??? 解:?sin111x1x?1,limx?0?e?0?limx2x?0esinx2?0又?limarcsinxxx?0x?limx?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?0limx2?x2sinnx?x?0xn 7

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n?40,limxn?20?n?2,n?4, 故2n原式?lim?cost?sin2t?t?01t x?0x2n?3

三、计算题(每小题8分,共64分)

13.求limsin3x?2xx??sin2x?3x

sin3x?2解: 原式=limsinxx??2x

x?3?limsin3xx??x?0??1??sin3x?1,limx??x?0??

limsin2xx??x?0???sin2x?1,lim1?x??x?0??

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

解:原式

有理化

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

?limtanxx??x?12?1x12limx?0x?2

15.求lim?21xx????sinx?cos?x??

解:令

1x?t,当x??时,t?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2x2lim2??x?0?4 ?1292?3x?????注:原式

????lim?2sin2xx?0cos2x?cos3x?3sin3x ????49 ex17.求lim?e?x?2xx?0x?sinx

0?x 解: 原式

0limex?e?2x?01?cosx 000ex?e?x0xlime?e?xx?0sinxlimx?0cosx?2 8

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??1x????ex?a,x?018.设f???1?cosx且limx?0f?x???x,x?0存在,求a的值。

解:?xlim??0??e?1x?a???e???a?0?a?a

??x21?cosx2xlim?0?x?limx?0?x1?lim2??x?x?0?x??22 ?a??22 119.lim?sin3x1?3lnxx?0??

解: 原式

?lim3cosx00?换底法ln(sin3x)x?0?sin3xlim3ex?0?1?3lnx?ex

?elim3xlimx1x?0?3sinx?ex?0?3x?e3

20.求lim??2?1??x???x?xln??1?x???? 1解: x?t原式

lim?1ln?t?0??1?t???tt2? ?通分limt?ln?1?t?t?0t2

??0??0??1?1lim1?tt?02t ?lim1?t?111t?02t?t?1??limt?0t?1?2

四、证明题(共18分) 21.当x??时且

limx??u?x??0,limx??v?x???,

证明lim?vlimu?x?v?x??1?x?x???u?x????ex??

证:lim?v?1?x?x???u?x???

1?lim??1u?x??u?x??v?x?x???u?x???

?exlim??u?x??v?x?

证毕

22.当x?0时,证明以下四个差函数的等价无穷小。

1)tanx?sinx等价于x3(2?x?0?

2)tanx?x等价于x3(3?x?0?

3)x?sinx等价于x3(6?x?0?

(4)arcsinx?x等价于x36?x?0?

证:?1??limtanx?sinxx?0x3

29

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??0??0??limtanx?1?cosx?x?0x3

2?x2x?lim2x?0x3?1 2当x?0时,tanx?sinx?x32

?22??limtanx?xx?01?limsecx?1 3x?0x23x?limtan2xx2x?0x2?limx?0x2?1 当x?0时,tanx?x?x23

?3??limx?sinx1?cosxx?01?lim3x?01 6x2x212x2?limx?01?1 22x当x?0时,x?sinx?136x

?4??limarcsinx?xx?01

6x3

1?lim1?x2?1lim1?1?x2x?01?2x?01 2x2x21?x212?lim2xx?01?1 2x2?1当x?0时,arcsinx?x等价于136x 五、综合题(每小题10分,共20分) 23.求limx???3x?9x2?12x?1?

2解: 原式

有理化lim9x??9x2?2x?1?x??3x?9x2?2x?1

?lim?2x?1x??3x?9x2?2x?1

?2?1?limx?21x?????

3?9?213?33x?x2limx224. 已知?mx?81x?2x2??2?n?x?2n?5,求常

数m,n的值。

解:(1)∵原极限存在且

limx?2??x2??2?n?x?2n???0 ?limx?2?x2?mx?8??0,4?2m?8?0

2m?12,m?6

(2)?limx2?6x?8x?2x2??2?n?x?2n

10


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