例1求lm tanx x→>0x 0-0 解原式=lim (tan x) lim sec x =1 x→>0(x x→0 例2求mimx-3x+2 0 x→)1y 3 x2-x+1 3x2-3 解原式=lim 6x3 lim x13x2-2x-1x16x-2 2例 1 解 . tan lim0 x x x → 求 ( ) (tan ) lim0   = → x x x 原式 1 sec lim 2 0 x x → = = 1 . 例 2 解 . 1 3 2 lim 3 2 3 1 − − + − + → x x x x x x 求 3 2 1 3 3 lim 2 2 1 − − − = → x x x x 原式 6 2 6 lim1 − = → x x x . 23 = ) 00 ( ) 00 (