find the value of $\lim_{x\to0}\frac{e-(1+x)^{\frac{1}{x}}}{x}$ find the value of $$\lim_{x\to0}\frac{e-(1+x)^{\frac{1}{x}}}{x}$$I use hospital law and can't find answer

Without using l'Hospital (I tend to avoid it as much as possible, as it always looked like a heavy hammer to me):
$$
(1+x)^{1/x} = e^{\frac{1}{x}\ln(1+x)} = e^{\frac{1}{x}(x-\frac{x^2}{2}+o(x^2))} = e^{1-\frac{x}{2}+o(x))} = e(1+\frac{x}{2}+o(x)) = e-\frac{xe}{2}+o(x)
$$
using the Taylor expansions of $\ln(1+x)$ and $e^x$ when $x\to0$. Plugging it back, the expression becomes
$$
\frac{e-(e-\frac{xe}{2}+o(x))}{x} = \frac{\frac{xe}{2}+o(x)}{x} = \frac{e}{2}+o(1)
$$
when $x\to0$. Hence, the limit is $\frac{e}{2}$.
