<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] In the same spirit as other answers, consider $$A=\left(\frac {\tan (x) }{x} \right)^{\frac{1}{x^2}}$$ Take logarithms $$\log(A)=\frac{1}{x^2}\log\left(\frac {\tan (x) }{x} \right)$$ Now, consider Taylor series $$\tan(x)=x+\frac{x^3}{3}+\frac{2 x^5}{15}+O\left(x^7\right)$$ $$\frac {\tan (x) }{x}=1+\frac{x^2}{3}+\frac{2 x^4}{15}+O\left(x^6\right)$$ $$\log\left(\frac {\tan (x) }{x} \right)=\log\left(1+\frac{x^2}{3}+\frac{2 x^4}{15}+O\left(x^6\right)\right)$$  Now use $\log(1+y)=y-\frac{1}{2}y^2+O\left(y^3\right)$ and replace $y$ by $\frac{x^2}{3}+\frac{2 x^4}{15}$ to get $$\log(A)=\frac{1}{x^2} \left(\frac{x^2}{3}+\frac{7 x^4}{90}+O\left(x^5\right)\right)=\frac{1}{3}+\frac{7 x^2}{90}+O\left(x^3\right)$$ which shows the limit and also how it is approached. 
You can continue using $A=e^{\log(A)}$ and still using Taylor arrive to $$A=\sqrt[3]{e}+\frac{7}{90} \sqrt[3]{e} x^2+O\left(x^3\right)$$
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