例2求加me-1~x x->0 解令ex-1=L,即x=lm(1+u) 则当x→0时,有u→0, e ∴Iim =lim M=m x→>0 n→In(1+l) n(1+ ne limIn(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