Find $\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$ $\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$
I let, $x = \frac{1}{n}$, then as 
$\lim_{x \to 0} \frac{1}{x}[(1+x)^\frac{1}{x} - e] = \infty$
L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{1} = \frac{1}{0}(1+0)^{\frac{1}{0}-1} = \infty$
Again, if we apply L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{x+(1+x)}$. This is also going to $\infty$ as $x \to 0$. But, I know the answer is $\frac{-e}{2}$, and I am not even close. Can anyone please find me the mistakes here. Oh, I am supposed to use L'Hopital's rule. 

note that :in line 4 you missed something ,that is $$y=(1+x)^{\frac{1}{x}} ,y' =?\\\ln y=\ln(1+x)^{\frac{1}{x}} \\\ln y=\frac{1}{x} \ln(1+x)\\\frac{y'}{y}=\frac{-1}{x^2}\ln(1+x)+\frac{1}{x} \frac{1}{1+x}\\y'= (1+x)^{\frac{1}{x}}\cdot \bigg(\frac{-1}{x^2}\ln(1+x)+\frac{1}{x} \frac{1}{1+x}\bigg)$$
