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?

Whenever we have an expression where both base and exponent are variables, it is best to take logs. Thus if $L$ is the desired limit then \begin{align} \log L &= \log\left\{\lim_{x \to 0}\left(\frac{\tan x}{x}\right)^{1/x^{2}}\right\}\notag\\ &= \lim_{x \to 0}\log\left(\frac{\tan x}{x}\right)^{1/x^{2}}\text{ (via continuity of log)}\notag\\ &= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(\frac{\tan x}{x}\right)\notag\\ &= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\frac{\tan x - x}{x}\cdot\dfrac{\log\left(1 + \dfrac{\tan x - x}{x}\right)}{\dfrac{\tan x - x}{x}}\notag\\ &= \lim_{x \to 0}\frac{\tan x - x}{x^{3}}\cdot 1\notag\\ &= \lim_{x \to 0}\frac{\sec^{2} x - 1}{3x^{2}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{3}\lim_{x \to 0}\frac{\tan^{2}x}{x^{2}}\notag\\ &= \frac{1}{3}\notag \end{align} Hence $L = e^{1/3}$.

Put $$ y = \left({\tan(x)\over x}\right)^{1/x^2}.$$ Then $$\log(y) = {\log(\tan(x)) - \log(x)\over x^2}$$ Can you evaluate the RHS with the help of L'hospital?

L'Hospital's rule is not the alpha and omega of limits computation!

As $\tan x= x+\dfrac{x^3}3+o(x^3)$, $$\frac{\tan x}x=1+\frac{x^2}3+o(x^2),\enspace\text{hence}\enspace \frac 1{x^2}\ln\Bigl(\frac{\tan x}x\Bigr)=\frac 1{x^2}\ln\Bigl(1+\frac{x^2}3+o(x^2)\Bigr)=\frac 13+o(1)\to \frac 13$$ so that $$\lim_{x\to 0}\Bigl(\frac{\tan x}x\Bigr)^{\tfrac 1{x^2}}=\mathrm e^{\frac 13}.$$