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?
 A: 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,
$$-\frac{1}{2}\leftarrow \frac{(1+x)\ln\left(\frac{1}{1-\frac{1}{1+x}}\right)}{2\ln x}=\frac{1}{2\ln x}\sum_{n=1}^{\infty}\frac{1}{n{{(1+x)}^{n-1}}}\geq\frac{1}{\ln x}\sum_{n=1}^{\infty}\frac{x}{{{(1+x)}^{n}}-{{(1-x)}^{n}}}.$$
A: 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 $-1/2$.
A: Well, for $0 < x < 1$, we have,
$S = \sum_{n=1}^{\infty}\frac{x}{(1+x)^n - (1-x)^n} 
=  \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{(1+x)^{n-1} + \cdots + (1-x)^{n-1}} 
\geq (\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n(1+x)^{n-1}}
= \frac{1}{1+x}\sum_{n=1}^{\infty}\frac{1}{n(1+x)^n}).$
Using power series for natural logarithm,
$S = -(1+x)\ln{\frac{x}{1+x}}$.
Now, let the required limit be L.
$L \leq (\lim_{x\to0^{+}}-((1+x)\frac{1}{2} - \frac{(1+x)\ln{1+x}}{2\ln{x}})
= -\frac{1}{2})$.
Hence, the upper bound.
