Evaluating $\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}$ How to evaluate the following limit? $$\lim_{x \to 0}\frac{(1+x)^{1/x} - e}{x}.$$

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 $0^{\pm}$
