Limit using L'hospital and maybe another trick

Hi, I've been sitting on this more than an hour and still can't find the solution :(

$\displaystyle \displaystyle\lim_{x \to{\infty}}{\displaystyle{x(}(1+\frac{1}{x})^{x} - {e})}$

I put this in a function grapher and it looks like the result should be $\displaystyle {-}\frac{e}{2}$, but I don't understand why this is true or how do I get to this solution. I assume that L'hospital should be used and supposedly there's another trick?

Thanks in advance

Simplifying the derivation

Actually, you can greatly simplify the derivation if you put y=1/x in the limit.

so as $\displaystyle x\to \inf , \\ y\to 0 $

the equation becomes

$\displaystyle

\lim_{y\to0} \frac{(1+y)^{1/y}-e}{y} $

applying l'hospital's rule, we get

$\displaystyle \lim_{y\to0} (1+y)^{1/y} \frac{y-(1+y)ln(1+y)}{y^2} $ which is

$\displaystyle \lim_{y\to0} (1+y)^{1/y} \lim_{y\to0} \frac{y-(1+y)ln(1+y)}{y^2} $

Since the left hand side is e, we apply l'hospital again to the right side

$\displaystyle e \lim_{y\to0} \frac{-ln(1+y))}{2y} $

Applying l'hospital again, we get

$\displaystyle \frac{-e}{2} $