# Math Help - Need detailed solution

1. ## Need detailed solution

I've tried multiplying with ln then L'hospital, but couldnt get to the end.

2. Originally Posted by Nrt

I've tried multiplying with ln then L'hospital, but couldnt get to the end.
Note that

$\left(\frac{x-1}{x+1} \right)^{x+1} = \left(\frac{x+1-2}{x+1} \right)^{x+1}=\left(1+\frac{-2}{x+1} \right)^{x+1}$

let $u=x+1$
Then as $x \to \infty \implies u \to \infty$
so
$\lim_{x \to \infty}\left(\frac{x-1}{x+1} \right)^{x+1}=\lim_{u \to \infty}\left(1+\frac{-2}{u} \right)^{u}=e^{-2}$