<s>[INST] Limit of tan function This is a question from an old tutorial for a basic mathematical analysis module. 
Show that $$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = e^2$$
My tutor has already gone through this in class but I am still confused. Is there anything wrong with the following reasoning? 
Since $\frac{\pi}{4}+\frac{1}{n} \to \frac{\pi}{4}$ as $n \to \infty$, 
it seems to me that $\tan(\frac{\pi}{4}+\frac{1}{n}) \to \tan(\frac{\pi}{4}) = {1}$,
and hence $\tan^n(\frac{\pi}{4}+\frac{1}{n}) \to 1^n = 1$.
Additionally, my tutor has given a hint, to use Squeeze theorem along with the definition $e = \lim\limits_{n \to \infty }(1 + {1\over n})^n$, but I can't see how these are to be used.
Edit: Here's another attempt I've made.
After using addition formula for tangent, we get 
$$\frac{1 + \tan{\frac{1}{n}}}{1- \tan{\frac{1}{n}}} = 1 - \frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = 1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}} = [(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}}$$
so given that $\lim\limits_{n \to \infty}{\frac{\tan{\frac{1}{n}}}{\frac{1}{n}}}=1$, we have 
$$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = \lim\limits_{n \to \infty}[(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2n\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = e^\frac{2*1}{1-0} = e^2$$
I'm really hoping that this method works as well! So sorry for the ugly formatting, I couldn't figure out some parts. 
 [/INST] If the only standard limit available is $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = e$$ then we are out of luck here. It is better to assume the following standard limits $$\lim_{x \to 0}\frac{\tan x}{x} = 1 = \lim_{x \to 0}\frac{\log(1 + x)}{x}$$ and then we can easily evaluate the desired limit. We have
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\tan^{n}\left(\frac{\pi}{4} + \frac{1}{n}\right)\right\}\notag\\
&= \log\left\{\lim_{n \to \infty}\left(\frac{1 + \tan(1/n)}{1 - \tan(1/n)}\right)^{n}\right\}\notag\\
&= \log\left\{\lim_{n \to \infty}\left(1 + \frac{2\tan(1/n)}{1 - \tan(1/n)}\right)^{n}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(1 + \frac{2\tan(1/n)}{1 - \tan(1/n)}\right)^{n}\text{ (via continuity of log)}\notag\\
&= \lim_{n \to \infty}n\log\left(1 + \frac{2\tan(1/n)}{1 - \tan(1/n)}\right)\notag\\
&= \lim_{n \to \infty}n\cdot\dfrac{2\tan(1/n)}{1 - \tan(1/n)}\cdot\dfrac{\log\left(1 + \dfrac{2\tan(1/n)}{1 - \tan(1/n)}\right)}{\dfrac{2\tan(1/n)}{1 - \tan(1/n)}}\notag\\
&= 2\lim_{n \to \infty}\dfrac{\tan(1/n)}{1/n}\cdot\dfrac{1}{1 - \tan(1/n)}\cdot\lim_{x \to 0}\dfrac{\log\left(1 + x\right)}{x}\notag\\
&= 2\cdot 1\cdot 1\cdot 1\notag\\
&= 2\notag
\end{align}
Hence $L = e^{2}$.
</s>