$\lim\limits_{x\to \infty} x\big(\log(x+1) - \log(x-1)\big) =e^2$ need to find the value of 
$$\lim\limits_{x\to \infty} x(\log(x+1) - \log(x-1))$$
$x(\log(x+1) - \log(x-1))=x(\log{ x+1\over x-1}) = \log({x+1\over x-1})^x = \log(1+{2 \over x-1})^x = \log(1+{2 \over x-1})^{x-1} + \log(1+{2 \over x-1})$
If we take $\lim\limits_{x \to \infty}$ at the last term,  $\lim\limits_{x \to \infty}\log(1+{2 \over x-1})^{x-1} = e^2, \lim\limits_{x \to \infty}\log(1+{2 \over x-1})=0$
Thus the answer is $e^2$
Is the above reasoning correct?

$$\lim_{x\to +\infty} x \log\frac{x+1}{x-1} = \lim_{x\to +\infty} 2x\,\text{arctanh}\left(\frac{1}{x}\right) = \lim_{z\to 0}\frac{2\,\text{arctanh } z}{z}=\lim_{w\to 0}\frac{2w}{\tanh(w)}$$
and since $\lim_{w\to 0}\frac{e^w-1}{w}=1$, the given limit equals $\color{red}{\large 2}$.
