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.
 A: 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}
A: let $$y = (1 + x)^{1/x} ,\quad \ln y = \frac 1 x\ln(1 + x) = \frac 1x\left(x - \frac 12x^2 + \cdots\right) = 1 - \frac x 2 + \cdots $$ so that we have $$y = ee^{-x/2+\cdots} = e\left(1 - \frac x 2 + \cdots\right) \to e - y = \frac{ex}2+\cdots$$
therefore $$ \frac{e-y}{\tan x} = \frac{ex/2 + \cdots}{x + \cdots} \to \frac e2 \text{ as } x \to 0.$$
finally, we have $$\lim_{x \to 0}\frac{e-(1+x)^{1/x}}{\tan x} = \frac  e2.$$
A: Since $e - (1+x)^{1/x} = \frac{xe}{2} + O(x)$ as $x \to 0,$ so
$$\frac{e - (1+x)^{1/x}}{\tan x} = \frac{\frac{xe}{2} + O(x)}{\sin x / \cos x} = \frac{\cos x \cdot [e/2 + O(x)]}{\sin x / x} \to \frac{e}{2}.$$
A: This is more a comment than an answer since previous answers show how to obtain the limit.
In the same spirit as abel's anwer, let me use a few more terms (keeping abel's notation) $$\ln y = \frac 1 x\ln(1 + x) =1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right) $$ $$y=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$ $$e-y=\frac{e x}{2}-\frac{11 e x^2}{24}+O\left(x^3\right)$$ $$\tan(x)=x+\frac{x^3}{3}+O\left(x^4\right)$$ $$\dfrac{e-\left(1 + x\right)^{\frac{1}{x}}}{\tan x}=\frac{\frac{e x}{2}-\frac{11 e x^2}{24}+O\left(x^3\right) }{x+\frac{x^3}{3}+O\left(x^4\right) }\approx \frac{\frac{e }{2}-\frac{11 e x}{24} }{1+\frac{x^2}{3} }\approx (\frac{e }{2}-\frac{11 e x}{24})(1-\frac{x^2}{3})$$  This will give $$\dfrac{e-\left(1 + x\right)^{\frac{1}{x}}}{\tan x}=\frac{e}{2}-\frac{11 e x}{24}+O\left(x^2\right)$$ which shows the limit but also the manner it is approached.
Edit
Just as a side result, suppose that you need to solve for $x$ the equation $$\dfrac{e-\left(1 + x\right)^{\frac{1}{x}}}{\tan x}=1$$ the given first order expansion gives an extimate of the solution $$x\approx\frac{12 (e-2)}{11 e}\approx 0.288$$ while the exact solution is $\approx 0.339$.
