Question: Any ideas on how to tackle this limit?
$$\lim\_{x \to 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}$$

I tried many ways but only got more complex stages, not easier ones...

Answer: 
By far the easiest and fastest way:
$$\lim\_{x \to 0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}} = \lim\_{x \to 0}\exp\left(\frac1{x^2}\ln\left(1 + \frac{x^2}{3}\right)\right) = \lim\_{x \to 0}\exp\left(\frac1{x^2}\cdot\frac{x^2}3\right) = e^{1/3}$$

Using the fact that, for $x \to 0$,
$$\begin{align}
\tan x &\sim x + \frac{x^3}3\\
\ln(1 + x) &\sim x
\end{align}$$
First, subtracting $(10)$ from $(9)$ in [this answer](https://math.stackexchange.com/a/438121), we get that
$$
\lim\_{x\to0}\frac{\tan(x)-x}{x^3}=\frac13\tag{1}
$$
Next, since $\lim\limits\_{x\to0}\left(1+tx^2\right)^{1/x^2}=e^t$ converges uniformly on compact sets,
$$
\begin{align}
\lim\_{x\to0}\left(\frac{\tan(x)}x\right)^{1/x^2}
&=\lim\_{x\to0}\left(1+\frac{\tan(x)-x}{x^3}x^2\right)^{1/x^2}\\
&=\lim\_{x\to0}\left(1+\frac13x^2\right)^{1/x^2}\\[6pt]
&=e^{1/3}\tag{2}
\end{align}
$$