Evaluate $\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$ Somebody asked this and I think it's quite interesting as I couldn't figure out how to evaluate this but the Wolfram Alpha says its limit is $\frac e2$.
$$\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$$
Could someone help here?
This limit can be evaluated by applying l'Hospital's rule twice. For the first time we differentiate 

$$(1+x)^{\frac{1}{x}}=e^{\frac{\ln (1+x)}{x}}$$ to get 

$$\frac{\frac{x}{1+x}-\ln (1+x)}{x^2}
e^{\frac{\ln (1+x)}{x}}.$$ 

Now $$e^{\frac{\ln (1+x)}{x}}\rightarrow e$$ so we need the limit of 
$$\frac{\frac{x}{1+x}-\ln (1+x)}{x^2}$$ another application of l'Hosptial gives

$$\frac{\frac{1}{(1+x)^2}-\frac{1}{1+x}}{2x}=-\frac{1}{2}\frac{1}{(1+x)^2}\rightarrow -\frac{1}{2}$$ 

So the limit is $-\frac{e}{2}$.
You can think of the limit  $$\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$$
as the derivative of the function $f(x)=(1+x)^\frac1x$ at point $x=0$.

Did that idea gave you any help?

Edit: You should define the value of $f(x)$ at $x=0$ namely : $f(0)=e$