TITLE: Evaluate $\lim_{x\to0}\frac{e-(1+x)^\frac1x}{x}$
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{e - (1 + x)^{1/x}}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\exp(1) - \exp\left(\dfrac{\log(1 + x)}{x}\right)}{x}\notag\\
&= -e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - 1}{x}\notag\\
&= -e\lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - 1}{\dfrac{\log(1 + x)}{x} - 1}\cdot\dfrac{\dfrac{\log(1 + x)}{x} - 1}{x}\notag\\
&= -e\lim_{x \to 0}\frac{\log(1 + x) - x}{x^{2}}\notag
\end{align}
The last limit can be easily evaluated either by Taylor series, or via L'Hospital's Rule or using integration and the value of this limit is $-1/2$ so that the answer to our question is $e/2$.