# Need detailed solution

• October 24th 2009, 12:01 PM
Nrt
Need detailed solution
http://www1.wolframalpha.com/Calcula...image/gif&s=55

I've tried multiplying with ln then L'hospital, but couldnt get to the end.
• October 24th 2009, 12:09 PM
TheEmptySet
Quote:

Originally Posted by Nrt
http://www1.wolframalpha.com/Calcula...image/gif&s=55

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