Finding a limit involving exponential function 
Find the value of the limit $$\lim_{x\to0}\left(\frac{e-(1+x)^\frac{1}{x}}{x}\right).$$

I tried to apply the standard limit $$\lim_{x\to0}(1+x)^\frac{1}{x} = e$$ and L'Hospital's Theorem individually, but that didn't help me. 
Any help will be appreciated.
 A: Using the standard Taylor expansions (to low order) $\ln(1+u)=u-\frac{u^2}{2}+o(u^2)$ and $e^u=1+u+o(u)$ when $u\to0$.
Rewriting 
$$\begin{align}
(1+x)^{\frac{1}{x}}
&= \exp\left( \frac{1}{x}\ln(1+x)\right)
= \exp\left( \frac{1}{x}(x-\frac{x^2}{2} + o(x^2))\right)
= \exp\left( 1-\frac{x}{2} + o(x)\right)
\\&= e\cdot \exp\left( -\frac{x}{2} + o(x)\right)
= e\cdot \left( 1-\frac{x}{2} + o(x)\right)
= e-e\frac{x}{2} + o(x)
\end{align}$$
we get that
$$
\frac{e-(1+x)^{\frac{1}{x}}}{x} = \frac{e\frac{x}{2} + o(x)}{x}
=\frac{e}{2} + o(1)\xrightarrow[x\to0]{} \boxed{\frac{e}{2}}.
$$
Note that we expanded $\ln(1+x)$ to order $x^2$, since we can "guess" the first order will get cancelled eventually by the $-e$ in the denominator. (Doing only the expansion to first order will  only, basically, yield the limit $(1+x)^{1/x}\xrightarrow[x\to0]{}e$, and so we know we need better precision.)
In the second step, factoring the $e$ out of the product allows us to get $e^{-x/2+o(x)}$ instead of $e^{1-x/2+o(x)}$, which is required in order to use the expansion of $e^u$ (since this expansion holds when $u\to 0$, and while $\frac{x}{2}\to 0$, this is not the case for $1-\frac{x}{2}$.)
A: Hint. Expand the function $f(x):=(1+x)^{\frac{1}{x}}=\exp\left(\frac{\ln(1+x)}{x}\right)$ at $0$:
$$f(x)=\exp\left(\frac{x-\frac{x^2}{2}+o(x^2)}{x}\right)=\exp\left(1-\frac{x}{2}+o(x)\right)=e\cdot \exp\left(-\frac{x}{2}+o(x)\right).$$
Can you take it from here? At the end you should find that the limit is $e/2$.
A: Put $$f (x)=(1+x)^\frac 1x $$ for $x\ne 0$ and $f (0)=e $.
then your limit is
$$-\lim_0\frac {f (x)-f (0)}{x-0} $$
or
$$-\lim_0 f'(x ).$$
with
$f'(x)=f (x)(-\frac {1}{x^2}\ln (1+x)+\frac {1}{x (1+x)}) $
$=f (x)(\frac {-x+x^2/2 (1+\epsilon (x))}{x^2}+\frac {1}{x}-\frac {1}{1+x}) $
from this, the limit is $$-e(\frac {1}{2}-1)=\frac {e}{2} $$
