Evaluation of $\bigg[ \lim_{x\to 0} \frac{e-(1+x)^{1/x}}{\tan x}\bigg]$ Find the value of 
$$\bigg[ \lim_{x\to 0} \frac{e-(1+x)^{1/x}}{\tan x}\bigg]$$
where $[.]$ represents floor function (or greatest integer function)
I wrote it as $\bigg[ \lim_{x\to 0} \frac{e-(1+x)^{1/x}}{x}\bigg]$ keeping in mind that $\lim_{x\to 0} \frac{tanx}{x}$ is slightly greater than $1$ but I cannot proceed further. Could someone suggest something. Given answer is 1.

In a neighbourhood of the origin,
$$(1+x)^{\frac{1}{x}}=\exp\left(\frac{\log(1+x)}{x}\right) = \exp\left(1-\frac{x}{2}+o(x)\right)=e\cdot\left(1-\frac{x}{2}+o(x)\right) $$
$$ \tan(x)=x+o(x) $$
hence the given limit equals the floor function applied to:
$$ \lim_{x\to 0}\frac{e-e\cdot\left(1-\frac{x}{2}\right)}{x}=\color{red}{\frac{e}{2}}.$$
Since $\frac e2$ is only slightly greater than 1, the floor of it is 1.
