<s>[INST] How to Find $ \lim\limits_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$. Can someone help me with this limit? I'm working on it for hours and cant figure it out.
$$ \lim_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$$
I started transforming to the form
$ \lim_{x\to 0} e^{  {\frac{\ln \left(\frac {\tan x}{x} \right)}{x^2}}  }$
and applied the l'Hopital rule (since indeterminated $\frac00$), getting:
$$ \lim_{x\to 0} \left( \frac{2x-\sin 2x }{2x^2\sin 2x} \right)$$
From here, I try continue with various forms of trigonometric substitutions, appling the l'Hopital rule again and again, but no luck for me. Can someone help me?
 [/INST] Let $\displaystyle y(x)
=\ln\left[\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}\right]
=\frac{1}{x^2}\ln\left(\frac{\tan x}{x}\right)$, then by using the 
L'Hôpital's rule several times,
\begin{align}
\lim_{x\to 0}y(x)&=\lim_{x\to 0}\frac{\ln\left(\frac{\tan x}{x}\right)}{x^2}\\
&\stackrel{{\rm H}}{=}
\lim_{x\to 0}\frac{\frac{x}{\tan x}\cdot\frac{x\sec^2x-\tan x}{x^2}}{2x}\tag{1}\\
&=\lim_{x\to 0}\frac{x\sec^2x-\tan x}{2x^2\tan x}\\
&\stackrel{{\rm H}}{=}
\lim_{x\to 0}\frac{\sec^2x+2x\sec^2x\tan x-\sec^2 x}{4x\tan x+2x^2\sec^2x}\\
&=\lim_{x\to 0}\frac{2\sec^2x\tan x}{4\tan x+2x\sec^2x}\\
&=\lim_{x\to 0}\frac{2\sec^2x}{4+2\left(\frac{x}{\sin x}\right)\sec x}\\
&=\frac{2\lim_{x\to 0}\sec^2x}{4+2\left(\lim_{x\to 0}\frac{x}{\sin x}\right)\left(\lim_{x\to 0}\sec x\right)}\\
&=\frac{1}{3}.
\end{align}
Hence $\displaystyle
\lim_{x\to0}\left(\frac{\tan x}{x}\right)^{\frac{1}{x^2}}
=\lim_{x\to0}e^{y(x)}
=e^{\lim_{x\to0}y(x)}=e^{\frac{1}{3}}$. Notice that $(1)$ holds because
$$\lim_{x\to0}\ln\left(\frac{\tan x}{x}\right)
=\ln\left(\lim_{x\to 0}\frac{\tan x}{x}\right)
\stackrel{{\rm H}}{=}
\ln\left(\lim_{x\to 0}\frac{\sec^2 x}{1}\right)=\ln(1)=0,$$
which follows that $\displaystyle \frac{\ln\left(\frac{\tan x}{x}\right)}{x^2}\rightarrow\frac{0}{0}$ and we can use the L'Hôpital's rule.
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