⑩天掌 Teaching Plan on Advanced Mathematics tanx SInx 解lim =lim x→0x x→ x cos x sInx →0xx→0cosy 1-cos x 例12求lm x→0 2 2 sin SIn 解原式=lm 2 =-lim x→>0 2x->0 SIn 2 2x→0x 2 2
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 1 cos 1 lim sin lim cos sin 1 lim tan lim 0 0 0 0 = = = → → → → x x x x x x x x x x x x 解 例12 求 . 1 cos lim 2 0 x x x − → 解 原式 2 2 0 2 2sin lim x x x→ = 2 2 0 ) 2 ( 2 sin lim 2 1 x x x→ = 2 0 ) 2 2 sin lim( 2 1 x x x→ = 2 1 2 1 = . 2 1 =
⑩天掌 Teaching Plan on Advanced Mathematics arcsin x 例13求lm x→0 解令t=sinx,则x=sin,当x→Q时,有t→>0 于是由复合函数的极算法则得 arcsin m x→>0 x→0smnt 2.单调有界准则 如果数列x满足条件 x1≤x2…≤xn≤xn1≤…,单调增加 单调数列 x1≥x2…≥xn≥xn+1≥…,单调减少 准则Ⅱ单调有界数列必有极限
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例13 求 x x x arcsin lim →0 解 于是由复合函数的极限运算法则得 令t = sin x,则x = sint,当x → 0时,有t → 0. 1 sin lim arcsin lim 0 0 = = → → t t x x x x 2.单调有界准则 如果数列x n满足条件 , x1 x2 xn xn+1 单调增加 , x1 x2 xn xn+1 单调数列 单调减少 准则Ⅱ 单调有界数列必有极限