第一章 一元函数的极限(4)

例

?x1?xx?8求极限lim???xx???e??x?1??

解 原极限可化简为

?x1?xx?lim????limxx???x???e??x?1??x?1???x?e??1???x??????. (*)

1??e?1??x??x(*)式中分母的极限为e2,因此只要求分子的极限.

x?1???limx?e??1????limx???x???x??????1??e??1??x??1xx?limy?0e??1?y?yy1(利用代换y?1x)

1?lim(1?y)y?0y?1?1??ln1?y??2?y?1?y???y (L’Hospital法则)

因此只要求

?1??1?y?ln?1?y??y?1?y?ln?1?y??y1lim????lim?limln1?y??222y?0y?0y?0y?1?y??y?1?y?y?y

(利用lim?1?y??1)?y?0limy?0ln?1?y??1?12y?12.

因此(*)式中分母的极限为

e2,最后得到

?x1?xx?2lim????xx???e?e??x?1?lim.

例9 设f(x)在?0,???上连续,且limf?x??a,求数列极限

x???n????10f?nx?dx．

解 将数列极限转化为函数极限,然后利用L’Hospital法则

limn????10f?nx?dx?limn????n0f?t?dtn (变量代换t?nx)?limy????y0f?t?dty(将n换成连续变量y)

?limy???f?y?(L’Hospital

法则) ?a

例10 求极限limx?0tanx?xsin3.

x3x解 利用Taylor公式 tanx?x?我们有

x??limx?03?ox??,sin3x?x?o?x?

x3limx?0tanx?xsin33?ox???33x1?limx?03?oxx??333x?x?o?x???13.

o?x????1??x??三、利用定积分求极限

定理2 (1)若f(x)在?a,b?上可积,则

?baf?x?dx?limn???b?ann?i?1b?a?b?a?f?a?i??limn???nn??1n?1?i?0b?a??f?a?i?n??.

(2) 若f(x)在?0,1?内单调,且积分?f?x?dx存在(可以是非正常积分),则

0?10f?x?dx?limn???1n?ni?1?i?f???n?(当0是瑕点时)?limn?????1n?1?ni?0?i?f???n? (当1是瑕点时).

(3) 若f(x)在?0,???内单调递减,且积分?0f?x?dx存在,则

???0f?x?dx?limh?f?nh?h?0n?0???.

证明 (1)由定积分定义直接得.

(2)当f(x)在?0,1?上可积时结论显然成立.

设?f?x?dx是非正常积分,不妨设0是瑕点并设f(x)在?0,1?内单调递减,显然有

0i?11?nin?i?f???f?x?dx?n?n?1i?nni?1nf?x?dx ?i?1,2,?,n?1?.

对i求和得 ?11f?x?dx?1nf?1??1n?i?1n?i?f????n??10f?x?dx.

令n???,得 ?01f?x?dx?limn???1n?ni?1?i?f???n?.注意到?11f?x?dx??1n0f?x?dx

（3）由于f(x)在?0,???内单调递减,对任意正数h有

??n?1?hnhf?x?dx?hf?nh???k?1?h??nhn?1?hf?x?dxk.?n?1,2,?,k?

对n求和得 ?hf?x?dx?h?f?nhn?1??????kh0f?x?dx.

令k???得 ?h??f?x?dx?h?f?nhn?1??????0f?x?dx.

再令h?0得 ?0

例11 求极限limn?????f?x?dx?limh?f?nh?h?0n?0?.

1k?2k???nk?1k,其中k是大于-1的实数.

n解 利用定理2得

limn???1k?2k???nk?1kn?limn???1n?ni?1?i????n?k??10xdx?k1k?1.

例12 设xnn??n?1??n?2???n?n?nnk,求极限limxn.

n???解 因为lnxn?2ln2?11n?i?1?i????n?,所以limlnxn?limn???1nnn????i?1i??ln?1???n???10ln?1?x?dx?2ln2?1,故

n???limxn?e?4e.

例13求极限limtann???1n?ln1nn!nn.

1n解 由于n???时, tan故原极限??limn???~.

limn???1n?ln?lnn!?nln1??limn???1n??ln1?lnn???ln2?lnn?????lnn?lnn??

1nn?i?1lnin??0lnxdx??1.

练习 (1) 求极限lim??n???n2?n?12?nn2?22???nn2?n22???.

.

(2) 求极限limn???1?22n?1?2?n?n2?22???n2??n?1???? 例14

? 求极限lim?n?????n?1n?2n?12n???n?1??2n??2.

解 原极限可变为

nlimn????k?1kn?1kn?limn???1n?nnk?1k2k?1??lim???n???nk?1?1nnk?1k2?1n???limn???nk??n?k?1nk.

1由于

1???nk?1??nnk?1k2?n???1k?n?n?k?1knk?1k2?1n2?nk?n1k2?0(n???),

k?1所以由定理2得 原式?limn???1n?nnk?k?1?1dxx?2.

0

例15 证明limt?1???1?t?tn?0n2??24.

分析 对于级数?n?0??tn2?1?t?t?t9?t16??,没有现成的求和公式可用.但是我们知道,当

t?1时级数收敛,当t??1时级数发散.又当t?1???时, ?n?0??tn2???.另一方面,注意到

?0e?x2dx??2,由定理2可得如下证明.

????2证明

?2??????0e?x2dx?limh?e?h?0n?0??nh?2?lim?t?1?lnt?tn?0n2(令t?e?h)

?lim?t?11?t?tn?0n2?lim?t?1lnt1?t???lim?t?11?t?tn?0n2

其中limt?1lnt??1是容易证明的.

1?t

例16 求极限limn???nn2?n!?.

解 令an?nn2?n!?,则limn???an?1an?limn????n?1?n?1?n!?2???n?1?!?2nn1???lim??1??n???n?1n??nn21n?0?1,所以级数

???n?1an收敛.从而由级数收敛的必要条件,liman?0,即limn???n????n!?n?0.

练习 求极限 (1) limn???5n?n!n?2n?,(2) limn???2n?n!nn,(3) limn???n3n.

?n!

例17 求极限limsinn?????n2?1??. ??0解 limsinn????nn2?1??limsinn???2n2?1??n??n???limn?????1?nsin?n2?1??n??

?limn?????1?sinn2?.

?1??n?2练习 求极限limsinn?????n2?n?.

.

23?45???2n2n?1

例18 求极限limn???1?3????2n?1?2?4????2n?2n?12n解 令xn0?xn??12?34???,

yn?,则xn?yn且xn2?yn?xn?12n?1.于是

12n?1?0(n???),即limxn?0.

n???

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