$\lim\limits_{x\to \infty} x\big(\log(x+1) - \log(x-1)\big) =e^2$ 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?
 A: $$x\log(x+1)-x\log(x-1)=x\log(1+\frac{1}{x})-x\log(1-\frac{1}{x})\underset{u=1/x}{=}\frac{\log(u+1)}{u}-\frac{\log(1-u)}{u}\underset{u\to 0}{\to} 2.$$
A: The reasoning is very correct...the answer is not: apparently you forgot you had logarithms and instead of $\;e^2\;$ the answer should be $\;\log e^2=2\;$ :
$$\lim_{x\to\infty}\color{red}\log\left(1+\frac2{x-1}\right)^x=\lim_{x\to\infty}\left[\color{red}\log\left(1+\frac2{x-1}\right)^{x-1}+\color{red}\log\left(1+\frac2{x-1}\right)\right]=$$
$$=\color{red}\log e^2+\color{red}\log1=\color{red}\log e^2+0=2$$
A: $$\lim_{x\to +\infty} x \log\frac{x+1}{x-1} = \lim_{x\to +\infty} 2x\,\text{arctanh}\left(\frac{1}{x}\right) = \lim_{z\to 0}\frac{2\,\text{arctanh } z}{z}=\lim_{w\to 0}\frac{2w}{\tanh(w)}$$
and since $\lim_{w\to 0}\frac{e^w-1}{w}=1$, the given limit equals $\color{red}{\large 2}$.
A: We have that as $x\to\infty$,
$$x\left(\log(x+1) - \log(x-1)\right)=\log\left(\frac{\left(1+\frac{1}{x}\right)^x}{\left(1+\frac{-1}{x}\right)^x}\right)\to\log\left(\frac{e}{e^{-1}}\right)=2$$
wher we used the limit $\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^x=e^a$.
