同除以sinx得1<x<1 SInx →c0SX< sInr cosx 5x 0<1-s<1-coS x=2sin2<2 →0 sInd 从而lim(1 x→0 +)=0→ lim sinx x→0 例16.求(1).lm tanx 0 tanx SIn 解.lim =lim x→>0x △→0 r cos cos x (2). lim x→0 x 2sint SIn SIn 解 lim lim llim--I x→0 2=lim x→0x x→0 2x-0x7 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 = = = =