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 2Tianjin 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 =