Evaluate a limit (probably involving L'Hôpital rule) Evaluate the limit:
$$\mathop {\lim }\limits_{x \to \infty } x\left( {{{\left( {1 + {1 \over x}} \right)}^x} - e} \right)$$
My attempts didn't yield a result. I'd be glad for a guidance.
Thanks!

You just have to persist in the L'Hopital rule. 
Start with
$$\frac{(1+\frac{1}{x})^x-e}{\frac{1}{x}}$$ top and bottom go to zero,
by L'Hopital we get
$$\frac{(\ln(1+\frac{1}{x})-\frac{1}{x+1})(1+\frac{1}{x})^x}{-\frac{1}{x^2}}
=(\frac{x^2}{x+1}-x^2\ln(1+\frac{1}{x}))(1+\frac{1}{x})^x$$
Now $(1+\frac{1}{x})^x\to e$ so lets set that term aside.
What remains can be written
$$\frac{\frac{1}{x+1}-\ln(1+\frac{1}{x})}{\frac{1}{x^2}}$$
top and bottom go to zero so again with L'Hopital,
$$\frac{-\frac{1}{(x+1)^2}+\frac{x}{x+1}\frac{1}{x^2}}{-\frac{2}{x^3}}=-\frac{1}{2}(\frac{x^2}{x+1}-\frac{x^3}{(x+1)^2})=-\frac{1}{2}\frac{x^2}{(x+1)^2}\rightarrow -\frac{1}{2}$$
So we get the limit $-\frac{e}{2}$
