Find the limit: $\lim_\limits{x\to 0}{\frac{\left(1+x\right)^{1/x}-e}{x}}$ 
Find the limit: $$\lim_\limits{x\to 0}{\frac{\left(1+x\right)^{1/x}-e}{x}}$$

I have no idea what to do, but I thought that this is the limit of the derivative of $f(x)=\left(1+x\right)^{1/x}$, as $x$ tends to 0. Any help?

If Taylor series (the "right" approach) are not yet available, let's use L'Hospital's Rule. 
When we differentiate the top we get 
$$(1+x)^{1/x}\left(\frac{x/(1+x)-\ln(1+x)}{x^2}\right).$$
The front part safely has limit $e$, so we only need to find
$$\lim_{x\to 0}\frac{x/(1+x)-\ln(1+x)}{x^2}.$$
One round of L'Hospital's Rule gets us to
$$\lim_{x\to 0}\frac{1/(1+x)^2-1/(1+x)}{2x}.$$
Now a little algebra finishes things. Or if one likes L'Hospital's Rule, do it again.
