How to compute $\lim _{x\to 0}\frac{(1+x)^{1\over x}-e}{x}$ without using a series expansion?

L'Hospital's Rule saves the day.

$$\begin{align} \lim_{x\to 0}\frac{(1+x)^{1/x}-e}{x}&=\lim_{x\to 0}\left(\frac{(1+x)^{1/x}}{x+1}\frac{x-(1+x)\log (1+x)}{x^2}\right)\\\\ &=e\lim_{x\to 0}\left(\frac{-\log (1+x)}{2x}\right)\\\\ &=-\frac{e}{2} \end{align}$$

While calculating limits if you find an expression of type $\{f(x)\}^{g(x)}$ (so that both base and exponent are variables rather than constants) then it is better to recast the expression in the form $\exp\{g(x)\log f(x)\}$ where $\exp(t)$ is just an equivalent notation for $e^{t}$. We can now proceed in the following manner \begin{align} L &= \lim_{x \to 0}\frac{(1 + x)^{1/x} - e}{x}\notag\\ &= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x}\right) - \exp(1)}{x}\notag\\ &= \lim_{x \to 0}\exp(1)\cdot\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_{t \to 0}\frac{e^{t} - 1}{t}\cdot\lim_{x \to 0}\frac{\log(1 + x) - x}{x^{2}}\text{ (putting }t = \left(\frac{\log(1 + x)}{x} - 1\right))\notag\\ &= e\cdot 1\cdot\lim_{x \to 0}\frac{\log(1 + x) - x}{x^{2}}\notag\\ &= e\lim_{x \to 0}\dfrac{\dfrac{1}{1 + x} - 1}{2x}\text{ (by L'Hospital's Rule)}\notag\\ &= -\frac{e}{2}\lim_{x \to 0}\frac{1}{1 + x}\notag\\ &= -\frac{e}{2} \end{align} It is always better to use L'Hospital's Rule (if needed) in conjunction with algebraic manipulation and some standard limits like $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1,\,\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$ This avoids repeated use of LHR as well as saves us from complicated expressions obtained via differentiation while applying LHR.