1- cosx例4 求 lim2x-→0xx-22 x2sin2 sin12lim解：原式=lim2x→0x-→0(2)x-2sinlim2xx-→0212
例 4 . 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
练习填空：元元cot一xsinox212(1) lim0 ; (5)limx→>11-xx-0x23sin2xtan3x(2) lim32(6) limsin3xx0x0sin2x1sinx(3)lim xcot3x = _3 _; (7)lim0x-→02xx-001x’ sinx0x(8) lim 2" sinx(4) lim2"n-00x→0sinx
0 (3)lim cot3 = _; x x x 0 sin (1)lim = _; x ωx x 0 sin2 (2)lim = _; x sin3 x x 练习 填空： ) ; 1 π cot 2 (5 lim = x 1 x x ω 2 3 1 3 π 2 ; 0 tan3 (6)lim = x sin2 x x 3 2 sin (7)lim = _; x 2 x x 0 (8)lim 2 sin . 2 n n n x x 2 0 1 sin (4) lim = x sin x x x 0