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个人收集整理 仅供参考学习
班级名称: 应用数学2班 学号: 200940510212 姓名:怀听听
探讨求极限地若干方法
引言:极限是数学中一项常用地“工具”,是学习数学必要掌握地方法之一,下面我们就来探讨一下求极限地几种方法:夹逼原理、常用极限法、等价无穷小量与无穷大量法则、洛比达法则、泰勒公式替代法、定积分法、连续性法.求极限有很多方法,还有有关级数方面地求法等,在此不作讨论.b5E2RGbCAP 1、夹逼原理求极限
夹逼原理:设数列?an?,?bn?,?cn?满足an?bn?cn,且liman?limcn?a,则
x??x??limbn?a
x??例题:求(1)lim11?n2?x??n??nn;
??11(2)lim???22x??n?2?n?1(3)lim1?3?5???2n?1?
x??2?4?6??2n1?1?1?n??? 2n?n?1解:(1)因为即1?lim?1??11?n2?n??nn??1?nnn?nn??nn,
?11?n2?n??nn?nn;而limnn?1;所以由夹逼原理得:
x???11?n2?x??n??nn?1
?(2)因为nnn?n2?1n?12?1n?22??1n?n2?nn?12?1,
而 limx???11?1,所以lim???222x??n?nn?2?n?11?3?5???2n?1?135????2?4?6??2n246?2n?224?un???2n?135?2n, 2n?1????1 2n?n?1 (3)设un?124则有???235?2n?1, 2n111112?un??un?将不等式同乘以un得?;即有 22n2n?12n2n?11 / 8
个人收集整理 仅供参考学习
而limx??12n?limx??1?0 2n?1因此lim1?3?5???2n?1??0
x??2?4?6??2n
2、常用极限法
sinx?1??1;常用极限:(1)lim(2)lim?1???e
x?0x??x?x?x例题:求(1)lim1?cosx???;(2) limcos??2x?0n??xn??22n2xx??2sinsin??1?cosx1122lim?lim?lim?解:(1) ??x?0x?0x2x22x?0?x?2?2??????????????(2)lim?cos??lim?1??cos?1???lim?1??cos?1??n??x??n??n?n??n???????又因为
22n2n2???n2?cos?1??n?cos?1?n1
??????2?sin?sin2??????22n??2?2nlimn?cos?1???lim??lim?; ???n??n??n??1?n22????2n?2n??所以lim?cos??en??n????n2??22
3、等价无穷法求极限 等价无穷小量:若lim作f?x?~g?x??x?x0?
当x?0时,常用等价无穷小:(ⅰ)ex?1~x;(ⅱ)ln?1?x?~x; (ⅲ)sinx~x;(ⅳ)1?cosx~f?x?g?x?x?x0?1,则称f与g是当x?x0时地等价无穷小量.记
12ax;(ⅴ)?1?x??1~ax 22 / 8
个人收集整理 仅供参考学习
例题:求(1)limx?0ln?1?3x3??e2x?1?sinx2;(2)limx?01?tanx?1?sinx 2xln?1?x??x??3 4解:(1)limx?0ln?1?3x3??e2x?1?sinx2?limx?0?3x3?2x?2x(2)
1?tanx?1?sinxlim?limx?0x?0xln?1?x??x2tanx?sinx?2?1?tanx?1?sinx?xln1?x?x??????limx?0?12x?xsinx?1?cosx?112lim??limx?0x?ln1?x?x? ??1?tanx?1?sinxcosxx?0x??ln?1?x??x??2???1x12x11?lim?lim??lim?1?x???4x?0ln?1?x??x4x?01?12x?021?x
4、洛比达法则求极限
洛比达法则:设:(1)当x?a时,函数f?x?及F?x?都趋于零;
(2)在点a地去心邻域内,f??及F??都存在且F???0; x?x?x? (3)当x?a时lim
x?a
200f??x?F??x?存在(或为无穷大),那么 x?a时limx?af?x?F?x??limx?af??x?F??x?.再设:(1)
当x??时,函数f?x?及F?x?都趋于零;
(2)当x?N时f??及F??都存在,且F???0; x?x?x?
(3)当x??时limx??f??x?F??x?存在(或为无穷大),那么
x??时
limx??f?x?F?x??limx??f??x?F??x?.p1EanqFDPw
例题:
?sinx?求(1)lim??x?0?x?11?cosx;(2)limxx x??3 / 8