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: 
Notice that

$$\tan x=x+\frac{x^3}3+o(x^3)$$
so
$$\left(\frac{\tan x}{x}\right)^{1/{x^2}}\sim\_0\left[\left(1+\frac{x^2}{3}\right)^{3/x^2}\right]^{1/3}\xrightarrow{x\to0}e^{1/3}\tag{$\because [1+1/x]^x\sim\_0e$}$$
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,
$$
\begin{align}
\lim\_{x\to0}\left(\frac{\tan(x)}x\right)^{1/x^2}
&=\lim\_{x\to0}\left(1+\frac{\tan(x)-x}{x^3}x^2\right)^{1/x^2}\\
&=\lim\_{x\to0}\left(1+\frac13x^2\right)^{1/x^2}\\[6pt]
&=e^{1/3}\tag{2}
\end{align}
$$