例2求mx-1 x→>0X 解令ex-1=u,即x=ln(1+u 则当x→0时,有u→>0, lim e< =lim L = im x→>0X a→0ln(1+u)"->0 In(1+u) 1 ne imln(1+u)“ →>0 即,当x→>0时,x~ln(1+x),x~ex-1.例２解 ln(1 ) lim 1 lim0 0 u u x e u x x + = − → →  . 1 lim0 x e x x − → 求 e 1 u, x 令 − = 即 x = ln( 1 + u), 则当 x → 0时,有 u → 0, u u u 1 0 ln( 1 ) 1 lim + = → u u u 1 0 limln( 1 ) 1 + = → ln e 1 = = 1 . → 0 ~ ln(1 + ), ~ − 1. x 即，当 x 时，x x x e