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?

It is possible to compute this limit only by making use of some basic limits:
\begin{eqnarray*}
\lim_{u\rightarrow 0}\frac{\ln (1+u)}{u} &=&1 \\
\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}} &=&\frac{1}{3} \\
\lim_{x\rightarrow 0}\frac{\tan x}{x} &=&1.
\end{eqnarray*}
Take the logarithm
\begin{eqnarray*}
L &=&\ln \left( \frac{\tan x}{x}\right) ^{1/x^{2}} \\
&=&\frac{1}{x^{2}}\ln \frac{\tan x}{x} \\
&=&\frac{1}{x^{2}}\ln \left( 1+\frac{\tan x}{x}-1\right)  \\
&=&\frac{\left[ \frac{\tan x}{x}-1\right] }{x^{2}}\cdot \frac{\ln \left( 1+%
\left[ \frac{\tan x}{x}-1\right] \right) }{\left[ \frac{\tan x}{x}-1\right] }
\\
&=&\frac{\tan x-x}{x^{3}}\cdot \frac{\ln (1+u(x))}{u(x)},\ with\ u(x)=\frac{%
\tan x}{x}-1
\end{eqnarray*}
since
\begin{equation*}
\lim_{x\rightarrow 0}u(x)=\lim_{x\rightarrow 0}(\frac{\tan x}{x}-1)=1-1=0
\end{equation*}
then
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\ln (1+u(x))}{u(x)}=\lim_{u\rightarrow 0}\frac{%
\ln (1+u)}{u}=1,
\end{equation*}
and therefore
\begin{equation*}
\lim_{x\rightarrow 0}L(x)=\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}\cdot
\lim_{x\rightarrow 0}\frac{\ln (1+u(x))}{u(x)}=\frac{1}{3}\cdot 1=1.
\end{equation*}
Backward, one take the exponential, and using continuity arguments, obtains
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{\tan x}{x}\right)
^{1/x^{2}}=\lim_{x\rightarrow 0}e^{L(x)}=e^{1/3}.
\end{equation*}
