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
 A: 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 $\frac{e}{2}$.
A: $$\lim_{x\to0}\frac{e-(1+x)^{\frac{1}{x}}}{x}\stackrel{l'H}=\lim_{x\to 0}\left(\frac1{x^2}\log(1+x)-\frac1{x(1+x)}\right)(1+x)^{1/x}=$$
$$=\lim_{x\to 0}\color{red}{(1+x)^{1/x}}\frac{(1+x)\log(1+x)-x}{x^2(1+x)}$$
Now, on the problematic factor above:
$$\lim_{x\to 0}\frac{(1+x)\log(1+x)-x}{x^2(1+x)}\stackrel{l'H}=\lim_{x\to0}\frac{\log(1+x)}{2x(1+x)+x^2}\stackrel{l'H}=$$
$$=\lim_{x\to0}\frac1{(1+x)\left(2(1+x)+4x\right)}=\frac1{1\cdot2}=\frac12$$
and together with the red factor above and arithmetic of limits, the limit is $\;\dfrac e2\;$ .
A: A little different (again no L'Hopital): Set $g(x) = [\ln (1+x)]/x = 1-x/2+x^2/3 - \cdots .$ Our expression can be written
$$-\frac{e^{g(x)}-e^{g(0)}}{x-0} \to -(e^g)'(0) = -e^{g(0)}g'(0)= -e\cdot (-1/2) = e/2.$$
