Evaluate the limit $\displaystyle\lim_{x \to 0}\frac{(e-\left(1 + x\right)^{1/x})}{\tan x}$. How to evaluate the following limit
$$\displaystyle\lim_{x\to 0} \dfrac{e-\left(1 + x\right)^{1/x}}{\tan x}$$
I have tried to solve it using L-Hospital's Rule, but it creates utter mess.
Thanks for your generous help in advance.

This is not possible to do without the use of differentiation (i.e. L'Hospital or Taylor) or some amount of integration to obtain the inequalities for logarithm function. The best approach seems to be L'Hospital's Rule.
We proceed as follows
\begin{align}
L &= \lim_{x \to 0}\frac{e - (1 + x)^{1/x}}{\tan x}\notag\\
&= \lim_{x \to 0}\dfrac{\exp(1) - \exp\left(\dfrac{\log(1 + x)}{x}\right)}{\tan x}\notag\\
&= \exp(1)\lim_{x \to 0}\dfrac{1 - \exp\left(\dfrac{\log(1 + x)}{x} - 1\right)}{\tan 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}\cdot\frac{x}{\tan x}\notag\\
&= -e\lim_{t \to 0}\frac{e^{t} - 1}{t}\cdot\lim_{x \to 0}\frac{\log(1 + x) - x}{x^{2}}\cdot\lim_{x \to 0}\frac{x}{\tan x}\text{ (putting }t = \frac{\log(1 + x)}{x} - 1)\notag\\
&= -e\cdot 1\cdot\lim_{x \to 0}\frac{\log(1 + x) - x}{x^{2}}\cdot 1\notag\\
&= -e\cdot\lim_{x \to 0}\dfrac{\dfrac{1}{1 + x} - 1}{2x}\text{ (via L'Hospital's Rule)}\notag\\
&= -\frac{e}{2}\lim_{x \to 0}\frac{-1}{1 + x}\notag\\
&= \frac{e}{2}
\end{align}
