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?
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