1.5两个重要极限 sIn 1. im =1 lim inu(x) x→>0x l(x)→>0 2.lim(1+-)=e im[1+-、J=e x→00 l(x)->∞ u(x) lim(1+u)"=e lim [1+w(x) w(x) = e →0 w(x)→>0
1.5 两个重要极限 1 sin lim 0 = → x x x 1. + = e → x x x ) 1 2. lim(1 1 ( ) sin ( ) lim ( ) 0 = → u x u x u x + = e → ( ) ( ) ] ( ) 1 lim [1 u x u x u x + = e → u u u 1 0 lim(1 ) + = e → ( ) 1 ( ) 0 lim [1 ( )]w x w x w x
sInx x->0 025 f(x)= 0.25
-15 -10 -5 5 10 15 -0.5 -0.25 0.25 0.5 0.75 1 x x f x sin ( ) = o 1 sin . lim 0 = → x x x 1
例1求lm SIn zx 3x size 2sin2x 2 解im lim x-0 3x 03·2t=nsim2x2 32x-102x3 例2求lim tan r x→少0y 解 tan x sinx 1 m x→0y x→>0cosx sInx =lim lim x→0xx→~0coSX
例2 . tan lim 0 x x x→ 求 解 x x x x x cos 1 lim sin lim →0 →0 = = 1. x x x x cos sin 1 lim 0 = x → x x tan lim →0 0 sin 2 lim . x 3 x → x 例1 求 解 x x x 3 sin2 lim →0 x x x 3 2 2sin2 lim 0 = → x x x 2 sin2 lim 3 2 2 →0 = . 3 2 =
例3求im 1-cOS x 》0 2 SIn 1-cos x 解 lim 2 x→>0 2 →>0 2 x SIn 21 SIn =-lim 2x→>0 =(im2)2 2x>0x 2 般:lim sinu(r) u(x)→>0 x
. 1 cos lim 2 0 x x x − → 求 2 2 0 ) 2 ( 2 sin lim 2 1 x x x→ = 2 2 0 2 0 2 2sin lim 1 cos lim x x x x x→ x→ = − 2 0 ) 2 2 sin (lim 2 1 x x x→ = 例3 解 . 2 1 = 1 ( ) sin ( ) lim ( ) 0 = → u x u x u x 一般：
例4求lm2xsin 3x 解mim2x·sin 3x 令t= 3x 2 sint 2 ==lim 2.. sint = lim 3t 3t3 3t
. 31 lim 2 sin x x x → 求 x x x 31 lim 2 sin → t t t sin 32 lim0 = → . 32 = 例 4 解 t x x t 3131 == == 令 t t t sin 31 lim 2 0 →