【参考借鉴】中国人民大学出版社(第四版)高等数学一第1章课后习题详解.doc 下载本文

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?lim2x22x?x?1?x?x?12x?1; 当x???时,

22x?x?1?x?x?12x??1; 当x???时,

22x?x?1?x?x?1故limx??x??,

?x2?x?1?x2?x?1?不存在

★2.计算下列极限:

(1)lim?1??n???111?1?2?3????n?1??2???n?;(2)lim; 2n??222?n33?n?1??n?2??n?3??n?2???2n?3?(3)lim;(4)lim。 n??n???n?1??2n?1??3n?2?5n3知识点:数列极限求法;

思路:(1)(2)需要先化简被求极限的式子,(3)(4)则利用有理分式极限的求法;

?1?1???1??11?2??2解:(1)lim?1??2???n??lim;

n??n??12??221?2(1?n?1)?n?1?11?2?3????n?1?2(2)lim?lim?; 2n??n??2nn2?n?1??n?2??n?3??1;

(3)limn??55n333?n?2???2n?3?93(4)lim??;

n???n?1??2n?1??3n?2?62★3.设

n?1?3x?2,x?0??f?x???x2?1,0?x?1,分别讨论x?0及x?1时f?x?的极限是否存在?

?21?x,??x知识点:分段点处函数的极限;左右极限; 思路:分段点函数的极限要左右极限分别求;

x?1)?1,lim(3解:当x?0时,lim(x?2)?2;故limf?x?不存在;

??2x?0x?0x?022x?1)?2,故limf?x??2; ?2,lim(x?1x?1?x?1x★4.已知limf?x??4及limg?x??1,limh?x??0,求:

当x?1时,lim?x?cg?x?h?x?;(2)lim;(3)lim?f?x??g?x??;(4)lim?f?x??h?x??

x?cf?x??g?x?x?cf?x?x?cx?cg?x?(5)lim x?ch?x?(1)limx?cx?c知识点:函数极限四则运算性质;

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思路:按性质求;

g?x?g?x?limx?c解:(1)lim??1;

4x?cf?x?limf?x?limh?x?h?x?x?c(2)lim??0;

x?cf?x??g?x?limf?x??limg?x?x?c?f?x??g?x???limf?x?limg?x??4;

x?cx?c(4)lim?f?x??h?x???limf?x?limh?x??0;

x?cx?cx?ch?x?g?x?h?x?limx?c??; (5)lim??0,而无穷小的倒数是无穷大,故

x?cg?x???hxlimg?x?x?c(3)limx?cx?cx?cx2?2x?k?4,求k的值; ★5.若limx?3x?3知识点:函数极限;

思路:分析求极限的过程,求出k的值;

x2?2x?kx2?3x?x?3?k?3k?3?lim?lim(x?1?) 解:limx?3x?3x?3x?3x?3x?3k?3lim(x?1?)?4,故必有k?3?0,即k??3; x?3x?32方法二:可由§1-8节无穷小比较来解:当x?3时,x?3?0;故此时必有x?2x?k?0, 故k??3;

?x2?1??★6.若lim??ax?b??0,求a,及b的值; x???x?1??知识点:同上;

?x?1??2x?ax?b?x?1?2x?ax?b x2?1?ax?b?解:

x?1x?1x?1x??1?a?x?2??1?b?,

x?1?x2?1?x?????ax?b?则由limlim[1?ax?2??1?b?]?0知,必有

?x??x???x?1x?1??21?a?0,?2?1?b?0,解得:a?1,b??1

习题1-7

★1.计算下列极限:

tan5xtanx?sinx1?cos2x;(2)limxcotx;(3)lim;(4)lim;

x?0x?0x?0x?0xxxsinxxsinx2arcsinxx?sinx(5)lim?;(6)lim;(7)lim;(8)lim;

x????xx?0x?0x?sinxx?03x1?cosx(1)lim知识点:两个重要极限;

思路:当函数用三角函数和幂函数表达时,可考虑变形成

的,可用下一节的等价无穷小代换来解更容易;

sin,其中

?0;但本题解法不是唯一

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tan5xsin5x1x(2)limxcotx?lim?lim??5?5;cosx?1;

x?0x?0x?0x?0sinxx5xcos5x?1?sinx??1?tanx?sinx?cosx??limsinx?lim?1?1??1?0?0

(3)lim?lim??x?0x?0x?0x?0xxx?cosx?1?cos2x2sin2xsinx?lim?lim2?2; (4)limx?0x?0x?0xsinxxsinxxxxx?lim??lim?2?2?2; (5)lim?x?0x?0x1?cosxx?02xsin2sin22sinxsin???t?sint(6)lim??x?tlim?lim?1;

x????xt?0t?0tt2arcsinx2t2(7)x?0?arcsinx?0,则limarcsinx?tlim?;

x?0t?03sint3x3sinxx1?1?limx?sinx1?1x?0sinxx?(8)lim?lim??0; x?0x?sinxx?0sinxx1?11?1?limx?0sinxx解:(1)lim★2.计算下列极限:

?1?x?(1)lim?1?x?;(2)lim?1?2x?;(3)lim??x?0x?0x???x?1x1x3x?1?;(4)lim?1??x???x?x1xkx?k?N?

?x?(5)lim??x??x?1??x?3?x?a?;(6)lim??x??x?a???0x;(7)limx?0?1?xe?;(8)lim11?xlnx?0x1?x知识点:重要极限:lim?1??1?1??e(或lim?1???e)

????思路:将函数表达式化成lim?1??0?n1?1??e(或lim?1???e),并利用指数函数运算性质

?????1(em?n?em?en,emn?em1x??)得出结果

11?(?1)?????x?x???lim?1???x????e?1; 解:(1)lim?1?x??lim??1???x??x?0x?0x?0????????11?2??x2x(2)lim?1?2x???lim?1?2x???e2;

x?0x?0????(3)lim??1?x??1??lim??1??x??x??x???x?kx3xx?3??1?x???lim?1????e3;

x?0???x?????x???k?3??1???1?1?????(4)lim?1???lim???x??x???x???x???e?k;

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?x?(5)lim??x??x?1??x?3?x?1?1??lim??x???x?1???1?x?x?3??1???lim?1??????x?????1?x?????1?x??x?3?x?1

???1????lim?1????x??1?x???????x????1?3/x?1?1/x?e?1;

x?a2ax?2ax?a2a??x?a???lim1?(6)lim??x????x??x?a???x?a?(7)lim?2a????lim?1??x???x?a???x?a2a????2a1?a/x?e2a;

x?0?1?xe?x1x?lim1?xex?0?x?1xex?ex?e1?e

1?x1?2x1?x11?x2x??1?x??ln?limln??limln1????x?0xx?01?xx?0?1?x??1?x??sinx??x,x?0?★★3.设f?x?1???2,x?0,求limf?x?。

x?0?x?1,x?0??(8)lim12x?lne?1;

知识点:分段函数的极限

求出

思路:可以先将f?x?化成f?x?1?或f?t?1?,以利用已知的函数表达式;或者,由已知f?x?1?,

f?x?的表达式,再求limf?x?。

x?0解:方法一:换元:limf?x?t?1?xx?0limf?t?1?,由已知

t?1limf?t?1??lim(?sint)??sin1,则limf?x???sin1;

t?1t?1x?0t方法二:令x?1?t,则x?t?1,代入已知得

?sin?t?1??sin?t?1??,???t?1?0t??1t?1t?1????f?t???2,t?1?0?f?t???2,t??1?t,?tt?1?0t??1???????sin?x?1?,??x??1x?1?sin?x?1???f?x???2,x??1,则limf?x??lim???sin1;

x?0x?0x?1?x,x??1???★4.已知lim??x?c???3,求c。

x??x?c??x2知识点:同题2 思路:同题2

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