<s>[INST] Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$ Find $\lim_{x\to0}\frac{(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}-e^2}{x^2}$

My attempt:
$\lim_{x\to0}(\tan(\frac{\pi}{4}+x))^{\frac{1}{x}}=e^2$
So the required limit is in $\frac{0}{0}$ form.
Then i used L hospital form.
$\lim_{x\to0}\frac{e^2(\frac{2}{x\cos 2x}-\frac{1}{x^2}\log(\frac{1+\tan x}{1-\tan x}))}{2x}$
I am stuck here.
 [/INST] Note that $$\tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 - \tan x}$$ and therefore
\begin{align}
L &= \lim_{x \to 0}\dfrac{\left(\tan\left(\dfrac{\pi}{4} + x\right)\right)^{1/x} - e^{2}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\left(\dfrac{1 + \tan x}{1 - \tan x}\right)^{1/x} - e^{2}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{1}{x}\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right)\right) - e^{2}}{x^{2}}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\exp\left(\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2\right) - 1}{x^{2}}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\exp\left(\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2\right) - 1}{\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2}\cdot\dfrac{\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2}{x^{2}}\notag\\
&= e^{2}\lim_{t \to 0}\dfrac{\exp(t) - 1}{t}\cdot\lim_{x \to 0}\dfrac{\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2}{x^{2}}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\dfrac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2}{x^{2}}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right) - 2x}{x^{3}}\notag\\
&= e^{2}\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right) - 2\tan x + 2\tan x - 2x}{x^{3}}\notag\\
&= e^{2}\left(\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right) - 2\tan x}{x^{3}} + 2\lim_{x \to 0}\frac{\tan x - x}{x^{3}}\right)\notag\\
&= e^{2}\left(\lim_{x \to 0}\dfrac{\log\left(\dfrac{1 + \tan x}{1 - \tan x}\right) - 2\tan x}{\tan^{3}x}\cdot\frac{\tan^{3}x}{x^{3}} + 2\lim_{x \to 0}\frac{\sec^{2} x - 1}{3x^{2}}\right)\text{ (via LHR)}\notag\\
&= e^{2}\left(\lim_{u \to 0}\dfrac{\log\left(\dfrac{1 + u}{1 - u}\right) - 2u}{u^{3}}\cdot 1 + \frac{2}{3}\lim_{x \to 0}\frac{\tan^{2}x}{x^{2}}\right)\text{ (putting }u = \tan x)\notag\\
&= e^{2}\left(\lim_{u \to 0}\dfrac{2\left(u + \dfrac{u^{3}}{3} + o(u^{3})\right) - 2u}{u^{3}}\cdot 1 + \frac{2}{3}\right)\notag\\
&= e^{2}\left(\frac{2}{3} + \frac{2}{3}\right)\notag\\
&= \frac{4e^{2}}{3}\notag
\end{align}
The simplifications done above are obvious and are aimed to simplify the use of series expansions and L'Hospital Rule. Also note that the variable $$t = \frac{1}{x}\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right) - 2 = \dfrac{\log\left(1 + \dfrac{2\tan x}{1 - \tan x}\right)}{\dfrac{2\tan x}{1 - \tan x}}\cdot\dfrac{\tan x}{x}\cdot\frac{2}{1 - \tan x} - 2$$ tends to $1\cdot1\cdot 2 - 2 = 0$ as $x \to 0$ and this fact has been used in the solution provided.
As a rule I prefer to use series expansions only when they are available via memory (like binomial, exponential, logarithmic and sine/cosine series) and try to use algebraic simplification so as to limit the use of series expansion to these simple series. If I need to do complicated stuff with series (like multiplying/dividing series or finding series of composite functions) I show the evaluation of series coefficients.
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