$\lim\limits_{x\to \infty} x\big(\log(x+1) - \log(x-1)\big) =e^2$ 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?

The reasoning is very correct...the answer is not: apparently you forgot you had logarithms and instead of $\;e^2\;$ the answer should be $\;\log e^2=2\;$ :
$$\lim_{x\to\infty}\color{red}\log\left(1+\frac2{x-1}\right)^x=\lim_{x\to\infty}\left[\color{red}\log\left(1+\frac2{x-1}\right)^{x-1}+\color{red}\log\left(1+\frac2{x-1}\right)\right]=$$
$$=\color{red}\log e^2+\color{red}\log1=\color{red}\log e^2+0=2$$
