《数学分析》课程教学资源（PPT课件）第三章 函数极限 第四节 两个重要极限

第四节两个重要极限

第四节 两个重要极限

C一、重要极限Bsin xlimx-0x4D元设单位圆O,圆心角ZAOB=x,(0<x<2作单位圆的切线，得△ACO扇形OAB的圆心角为x△OAB的高为BD于是有sinx=BD,，x=弧AB,tanx = AC

A C 1 sin lim 0 = → x x x ) 2 , , (0  设单位圆 O 圆心角AOB = x  x  于是有sin x = BD, x = 弧AB, tan x = AC, x o B D 作单位圆的切线，得ACO. 扇形OAB的圆心角为x, OAB的高为BD, 一、重要极限

sinx即 cos x<.. sinx <x<tanx,<1,x"时,上式对于_元当0<<<x<0也成立222tx"0<cosx-1=1-cosx = 2sin22?t= 0,.:. lim(1 - cos x) = 0,: lim2x-0x-→0 sin x又: lim1= 1,:. lim1:. limcos x = 1,x-→0x→0x→0x

sin x  x  tan x, 1, sin cos   x x 即 x 0 . 2 上式对于   也成立  − x , 2 当 0 时   x  0  cos x − 1 = 1 − cos x 2 2sin2 x = 2 ) 2 2( x  , 2 2 x = 0, 2 lim 2 0 = → x x  lim(1 cos ) 0, 0  − = → x x limcos 1, 0  = → x x lim1 1, 0 = x→ 又 1. sin lim 0  = → x x x

1 - cos x例1 求 lim1.2x→0x2 sin?xsin?t￥12 2解 原式=limlim.2X/2x→02 x→0x2°xsin21.1-!limx22 x-→022

例 1 . 1 cos lim 2 0 x x x − → 求 解 22 0 2 2sin lim x x x → 原式 = 2 2 0 ) 2 ( 2 sin lim 21 x x x → = 2 0 ) 2 2 sin lim ( 21 x x x → = 2 1 21 =  . 21 =

1lim(1 + =)*二、重要极限=ex→8x1设x, =(1+-)"nn(n-1)...(n-n+1) 111)1n(n-n.=1+++2!n!n"1! nn1n-11-2)..-(1--=1+1+2.n!nn1XntX n+121n-1=1-2.n+1n+ 2n+1nn+121n+(n +)n+n++

e x x x + = → ) 1 lim(1 n n n x ) 1 设 = (1 +  + − = +  + 2 1 2! 1 ( 1) 1! 1 n n n n n ). 1 ) (1 2 )(1 1 (1 ! 1 ) 1 (1 2! 1 1 1 n n n n n n − = + + − ++ − −  − n n n n n n n 1 ! ( 1) ( 1)  − − + +  ). 1 ) (1 2 2 )(1 1 1 (1 ( 1)! 1 ) 1 1 ) (1 2 2 )(1 1 1 (1 ! 1 ) 1 1 (1 2! 1 1 1 1 + − + − + − + + + − − + − + + + − + = + + − + n n n n n n n n n n n xn    二、重要极限

显然xn+1>Xn,，：{x}是单调递增的;-+:x..一+22.1(x,}是有界的;n

, 显然 xn+1  xn  是单调递增的;  x n ! 1 2! 1 1 1 n xn  + + ++ 1 2 1 2 1 1 1 −  + + + +  n 1 2 1 3 − = − n  3,  是有界的;  x n lim 存在. n n x →  e n n n + = → ) 1 记为lim(1 (e = 2.71828)

当x≥1时,有[x]≤x≤[x]+1,11)[x]+1)Ixl ≤(1+-)*≤(1+_(1 +[x][x] + 1x11[x]+1Y而 lim (1 +lim (1mm[x]X-→+00x-→+00x>+α[x][x]1[x]lim (1x→+0[x]+ 11[x/+11= lim(1 +lim(1 -e.[x] +1'x-→+00[x] + 1x→+80.:. lim (1 + =)*ex→+00x

当 x  1时, 有[x]  x  [x]+ 1, ) , [ ] 1 ) (1 1 ) (1 [ ] 1 1 (1 [ ] [ ]+1  +  + + + x x x x x x ) [ ] 1 ) lim (1 [ ] 1 ) lim (1 [ ] 1 lim (1 [ ] 1 [ ] x x x x x x x x + = +  + →+  →+  + →+  而 = e, [ ] 1 1 [ ] ) [ ] 1 1 ) lim (1 [ ] 1 1 lim (1 ) [ ] 1 1 lim (1 − →+  + →+  →+  +  + + = + + + x x x x x x x x = e, ) . 1 lim (1 e x x x  + = →+ 

令t =-x,.. im (1+ l)* - lim(1 - = lim(1 + _x→-8f→+8+8x= lim(1 +t→+8t-11lim(1+-)*=ex→0x11lim(1 + x)* = lim(1 + -)= -xx→0α1lim(1 + x)* = ex-

令 t = −x, t t x x x t − →−  →+   + = − ) 1 ) lim(1 1 lim (1 t t t ) 1 1 lim (1 − = + →+ ) 1 1 ) (1 1 1 lim(1 1 − + − = + − →+  t t t t = e. e x x x  + = → ) 1 lim(1 , 1 x 令 t = t t x x t x ) 1 lim(1 ) lim(1 1 0 + = + → → = e. x e x x + = → 1 0 lim(1 )

例2 求 lim(1-=)*x-→8X1解 原式= lim[(1 +)-xilim18X8x-X(1 +13求 lim(例3x-→82-+1解原式= lim[(1 +x+2X→00x+2

例 2 ) . 1 lim ( 1 x x x − →  求 解 x x x − →  − + = ) 1 (1 1 ) ] 1 lim 1 lim[(1 − − → − = + x x x 原式 . 1e = 例 3 ) . 23 lim( 2 x x xx ++ →  求 解 2 2 4 ) 2 1 ) ] (1 2 1 lim[(1 + − → + + + = + x x x x 原式 . 2 = e

三、小结1.两个准则夹逼准则；单调有界准则2.两个重要极限设α为某过程中的无穷小，sinα1° lim: 1:某过程α12° lim(1 + α)α = e.某过程

1.两个准则 2.两个重要极限 夹逼准则; 单调有界准则 . 1; sin 1 lim 0 =   某过程 2 lim (1 ) . 1 0 +  = e  某过程 设 为某过程中的无穷小 , 三、小结