How can I calculate $\lim\limits_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$? 
How can I calculate this limit?
  $$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$$  

I thought about L'Hospital because case of $\frac{0}{0}$, but I don't know how to contiune from this point..

let $n = \frac 1x$
$\lim_\limits{n\to \infty} n((1+\frac 1n)^n - e)$
Binomial expansion:
$n(1 + 1 + \frac 12 (1-\frac 1n) + \frac 1{3!} (1-\frac 1n)(1-\frac 2{n})+\cdots +\frac {1}{n!} (1-\frac 1n)\cdots(1-\frac {n-1}{n}) - e)$
$e = 1 + 1 + \frac 1{2!} + \frac {1}{3!}\cdots$
$n( (-\frac 12  -\frac 1{3!} {3\choose 2} - \frac 1{4!} {4\choose 2} -\cdots - - \frac 1{n!} {n\choose 2})\frac 1{n} + o(\frac 1{n^2}))\\
-\frac 12(1+1+\frac 12 + \frac 1{3!}\cdots \frac {1}{n!}) + o(\frac 1n)\\
-\frac 12 e$
