L'Hôpital's as x tends to infinityI'm searching for the explanation to the limit of: lim...

Hint:
$\ln \frac{x+1}{x-1}=\ln (x+1)-\ln (x-1)$
so that
$x\cdot \ln \frac{x+1}{x-1}$
can be rewritten as
$\frac{\ln (x+1)-\ln (x-1)}{x^{-1}}.$
Now,
$\underset{x\to \mathrm{\infty }}{lim}x\ln \frac{x+1}{x-1}=\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{1}{x+1}-\frac{1}{x-1}}{\frac{-1}{x^2}}$
Find a common denominator for the two terms in the numerator, cancel fractions, and you will arrive at the desired answer of 2.
Note: Indeed, $\ln \frac{x+1}{x-1}\to 0$ and $x^{-1}\to 0$ as $x\to \mathrm{\infty }$. Therefore, we have the required form $0/0$ and L'Hopital's rule is applicable.