0 例1求lim tanx 0C 解原式=lim (tan))=lim sec x→>0(x x→0 3x+2 例2求im x→)1x-x2-x+1 3x2-3 解原式=lim 6x x→13x2-2x-1x→16x-2 3-2 Economic-mathematics 27-12 Wednesday, February 24, 2021Economic-mathematics 27- 12 Wednesday, February 24, 2021 例1 解 . tan lim 0 x x x→ 求 ( ) (tan ) lim 0   = → x x x 原式 1 sec lim 2 0 x x→ = = 1. 例2 解 . 1 3 2 lim 3 2 3 1 − − + − + → x x x x x x 求 3 2 1 3 3 lim 2 2 1 − − − = → x x x x 原式 6 2 6 lim 1 − = → x x x . 2 3 = ) 0 0 ( ) 0 0 (