<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] 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*}
</s>