同除以sin得1 1→c0sx<smx<1 sInr cos x 故0<1- sInx <1-cosx=2sin2<2·()2=→>0 从而lim(1-x)=0→lim sInd x→0 x→)0 tan x 例16.求(1)lim 解limx= lim sinx.1 0C A→0 Cosx 1-cos x (2)im using SIn SIn cos x 解lim lim -lim lim x→0 x→0y 2x-0 2x->0x 27 从而 1 sin 1 cos 1 sin cos x x x x x x      2 sin 2 2 0 1 1 cos 2sin 2 ( ) 0 2 2 2 x x x x x x  −  − =   = → 0 0 sin sin lim(1 ) 0 lim 1 x x x x → → x x − =  = 例16. 求 0 tan (1).lim ; x x → x 同除以 sinx 得 故 2 0 1 cos (2).lim ; x x → x − 0 0 tan sin 1 . lim lim 1 x cos x x → → x x x 解 =  = 2 2 2 2 2 0 0 0 0 2 2sin sin sin 1 cos 1 1 1 2 2 2 lim lim lim [lim ] 2 2 2 ( ) 2 2 x x x x x x x x → → → → x x x x − 解 = = = =