Evaluating $\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}$ How to evaluate the following limit? $$\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}.$$
 A: You may write 
$$
\begin{align}
\frac{(1+x)^{1/x} - e}{x} &= \frac{\large e^{\large\frac{\log (1+x)}{x}} - e}{x}\\\\
&= \frac{e^{\large \frac{x-\frac{x^2}{2}+{\mathcal{O}}(x^3)}{x}} - e}{x}\\\\
&= \frac{ e^{1-\frac{x}{2}+{\mathcal{O}}(x^2)} - e}{x}\\\\
&= \frac{ e \:e^{-\frac{x}{2}+{\mathcal{O}}(x^2)} - e}{x}\\\\
&= \frac{ e \:\left(1-\frac{x}{2}+{\mathcal{O}}(x^2)\right)- e}{x}\\\\
&=-\frac{e}{2}+{\mathcal{O}}(x)
\end{align}
$$
giving

$$\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}=-\frac e2$$

where we have used
$$
\begin{align} &\log(1+x) =x-\frac{x^2}{2}+\frac{x^3}{3}+ \cdots, \quad |x|<1, \\
&e^x =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+ \cdots .
\end{align}
$$
A: Since you already received very good answers, let me make the problem slightly more difficult and ask how the expression tend to $-\frac e2$.
As already answered, start with $$\frac{(1+x)^{1/x} - e}{x} = \frac{e^{\log (1+x)/x} - e}{x}$$ Now, $$\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right)$$ $$\frac{\log(1+x)}{x}=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$ Now, define $$y=-\frac{x}{2}+\frac{x^2}{3}+\cdots$$ and $$e^{1+y}=e~ e^y=e\Big(1+y+\frac{y^2}{2}+\cdots\Big)$$ and replace $y$ by its definition from $x$. It not be too long to find that $$\frac{(1+x)^{1/x} - e}{x}=-\frac{e}{2}+\frac{11 e x}{24}+O\left(x^2\right)$$ which shows how the expression behaves when $x$ tend to $0^{\pm}$
