TITLE: Finding the $\lim\limits_{x\to 0} \frac{(1+\tan x)^{1/x} - e}{x} $
QUESTION [7 upvotes]: How do I find:
$$\lim\limits_{x\to 0} \frac{(1+\tan x)^{1/x} - e}{x} $$
I tried L'Hospital but it does can't be applied since it's not in an indeterminate form.
Can I have some assistance?  Thanks!

REPLY [10 votes]: Recall the Taylor expansions at $0$ of
$$
e^u=1+u+O(u^2)\qquad \ln(1+v)=v-\frac{v^2}{2}+O(v^3)
$$
and 
$$
\tan x=x+O(x^3).
$$
Thus
$$
\frac{\ln(1+\tan x)}{x}=\frac{1}{x}\left(\tan x-\frac{\tan^2x}{2}+O(\tan^3x)\right)
$$
$$
=\frac{1}{x}\left(x-\frac{x^2}{2}+O(x^3) \right)=1-\frac{x}{2}+O(x^2).
$$
Then
$$
(1+\tan x)^\frac{1}{x}=\exp \left(1-\frac{x}{2}+O(x^2)\right)=e\exp \left(-\frac{x}{2}+O(x^2)\right)
$$
$$
=e\left(1-\frac{x}{2}+O(x^2)\right)=e-\frac{e}{2}x+O(x^2).
$$
Finally, 
$$
\frac{(1+\tan x)^\frac{1}{x}-e}{x}=-\frac{e}{2}+O(x)\longrightarrow -\frac{e}{2}.
$$