<s>[INST] 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?
 [/INST] 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.
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