I've tried multiplying with ln then L'hospital, but couldnt get to the end.
Note that
$\displaystyle \left(\frac{x-1}{x+1} \right)^{x+1} = \left(\frac{x+1-2}{x+1} \right)^{x+1}=\left(1+\frac{-2}{x+1} \right)^{x+1}$
let $\displaystyle u=x+1$
Then as $\displaystyle x \to \infty \implies u \to \infty$
so
$\displaystyle \lim_{x \to \infty}\left(\frac{x-1}{x+1} \right)^{x+1}=\lim_{u \to \infty}\left(1+\frac{-2}{u} \right)^{u}=e^{-2}$