1-cosx 例1求lm22 2sin SIn 解原式=im 2 lim 2 →>0x 2x→>0x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 =