Find $\lim\limits_{x\rightarrow 0^+}\frac{1}{\ln x}\sum\limits_{n=1}^{\infty}\frac{x}{(1+x)^n-(1-x)^n}$ 
Find $$\lim\limits_{x\rightarrow 0^+}\dfrac{1}{\ln x}\sum_{n=1}^{\infty}\dfrac{x}{(1+x)^n-(1-x)^n}.$$

Consider $$f(x):=(1+x)^n,$$
By Lagrange MVT, we can obtain 
$$\frac{2x}{f(x)-f(-x)}=\frac{1}{f'(\xi)}, -x\gtrless \xi\gtrless x$$
Thus $$\frac{x}{(1+x)^n-(1-x)^n}=\frac{1}{2f'(\xi)}=\frac{1}{2n(1+\xi)^{n-1}}$$
Can we go on from here?

If $x>0$, then by the mean value argument
$$
\frac{x}{{(1 + x)^n  - (1 - x)^n }} > \frac{1}{{2n(1 + x)^{n - 1} }}.
$$
Assume now that $0<x<1$. Then
$$
\frac{1}{{\log x}}\sum\limits_{n = 1}^\infty  {\frac{x}{{(1 + x)^n  - (1 - x)^n }}}  < \frac{1}{{\log x}}\sum\limits_{n = 1}^\infty  {\frac{1}{{2n(1 + x)^{n - 1} }}} 
\\
 =  - \frac{{1 + x}}{{2\log x}}\log \left( {1 - \frac{1}{{1 + x}}} \right) =  - \frac{{1 + x}}{2} + \frac{{1 + x}}{{2\log x}}\log (1 + x).
$$
Consequently,
$$
\mathop {\lim }\limits_{x \to 0 + } \frac{1}{{\log x}}\sum\limits_{n = 1}^\infty  {\frac{x}{{(1 + x)^n  - (1 - x)^n }}}  \le  - \frac{1}{2}.
$$
This, together with Robert Z's answer, imply that the limit is $-1/2$.
