$\lim_{x\rightarrow 0} \frac{e-(1+x)^{\frac{1}{x}}}{\tan x}$ Fund the value of $$\lim_{x\rightarrow 0} \frac{e-(1+x)^{\frac{1}{x}}}{\tan x}$$
I have got the answer but by using series expansion  and answer is $\frac{e}{2}$, can this be solved by using L.Hospital rule. This question was forwarded to me by a student who dont want to use expansion.
 A: Note that 
$$\frac{e-(1+x)^{\frac{1}{x}}}{\tan x}=
e\cdot  \frac{1-\exp\left(\frac{\ln(1+x)}{x}-1\right)}{\left(\frac{\ln(1+x)}{x}-1\right)}\cdot \frac{\ln(1+x)-x}{x^2}\cdot \frac{x}{\tan(x)}.$$
Now recall that 
$$\lim_{x\to 0}\frac{\tan(x)}{x}=1,\quad \lim_{x\to 0}\frac{\ln(1+x)}{x}=1,\quad \lim_{t\to 0}\frac{e^t-1}{t}=1,$$
and use L'Hospital rule for the remaining factor,
$$\lim_{x\to 0}\frac{\ln(1+x)-x}{x^2}=
\lim_{x\to 0}\frac{1/(1+x)-1}{2x}=\lim_{x\to 0}\frac{-x}{2x(1+x)}=-\frac{1}{2}.$$
Hence, putting all together, we obtain
$$\lim_{x\to 0}\frac{e-(1+x)^{\frac{1}{x}}}{\tan x}=
e\cdot (-1)\cdot\left(-\frac{1}{2}\right)\cdot 1=\frac{e}{2}.$$
A: First of all,
$\lim_{x \to 0} \dfrac{\tan x}{x}
=1
$,
so the denominator
can be replaced by $x$.
Then,
as $x \to 0$,
$\log(1+x) 
= x-x^2/2+O(x^3)$
and
$e^x
= 1+x+O(x^2)$
so that
$\begin{array}\\
(1+x)^{1/x}
&=e^{\log(1+x)/x}\\
&=e^{(x-x^2/2+O(x^3))/x}\\
&=e^{1-x/2+O(x^2)}\\
&=e\cdot e^{-x/2}e^{O(x^2)}\\
&=e\cdot (1-x/2+O(x^2))(1+O(x^2))\\
&=e\cdot (1-x/2+O(x^2))\\
&=e-ex/2+O(x^2)\\
\text{so that}\\
e-(1+x)^{1/x}
&=ex/2+O(x^2)\\
\text{and}\\
\dfrac{e-(1+x)^{1/x}}{x}
&=e/2+O(x)\\
\end{array}
$
Note:
Wolfy agrees, saying
the quotient is
$e/2 - (11 e x)/24 + (7 e x^2)/16 - (2447 e x^3)/5760 + (959 e x^4)/2304 + O(x^5)
$.
A: The denominator can be replaced by $x$ (thanks to the limit of $\tan x/x$).
Then the derivative of the numerator is
$$(e-(1+x)^{1/x})'=-(1+x)^{1/x}\left(\frac{\log(1+x)}x\right)'$$ and the first factor tends to $e$.
Then
$$\frac{x-(1+x)\log(1+x)}{x^2(1+x)}$$ is also an indeterminate form, which we can evaluate by two applications of L'Hospital as
$$\frac{1-\log(1+x)-1}{2x+3x^2}\to-\frac{\dfrac1{1+x}}{2+6x}\to-\frac12.$$
