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