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?

The limit is equal to $-\dfrac{1}{2}$. 
A LOWER bound: for $0<x<1$, $\log(x)<0$ and
$$\begin{align}
\frac{1}{\ln x}\sum_{n=1}^{\infty}\frac{x}{(1+x)^n-(1-x)^n}
&\geq
\frac{1}{2\ln x}\sum_{n=1}^{\infty}\frac{1}{n+\binom{n}{3}x^2}\\
&\geq 
\frac{1}{2\ln x}\int_{1/2}^{\infty}\frac{ds}{s+\frac{s^3x^2}{6}}\\
&=\frac{\frac{1}{2}\ln(x^2+24)-\ln x}{2\ln x}\to -\frac{1}{2}
\end{align}$$
As regards the UPPER bound, for $0<x<1$, we have that  $0<\frac{1}{1+x}<1$ and, from your work,
$$-\frac{1}{2}\leftarrow \frac{(1+x)\ln\left(\frac{1}{1-\frac{1}{1+x}}\right)}{2\ln x}=\frac{1}{2\ln x}\sum_{n=1}^{\infty}\frac{1}{n{{(1+x)}^{n-1}}}\geq\frac{1}{\ln x}\sum_{n=1}^{\infty}\frac{x}{{{(1+x)}^{n}}-{{(1-x)}^{n}}}.$$
