lim(x→0) [ln√（1+x/1-x)] / x
=lim(x→0) (1/2x)*ln[(1+x)/(1-x)]
=1/2 lim(x→0) [ln(1+x)-ln(1-x)] / x
(因为x→0时,ln(1+x)→0、ln(1-x)→0 、 x→0,上下同时求导)
=1/2 lim(x→0) [ln(1+x)]'/x' -1/2 lim(x→0) [ln(1-x)]'/x'
=1/2 lim(x→0) 1/(1+x) -1/2 lim(x→0) [-1/(1-x)]
=1/2 [1/(1+0)] + 1/2 [1/(1-0)]
=1/2 + 1/2
=1
所以,当x→0时无穷小量ln√（1+x/1-x）与x是等价无穷小