On the way finding limit. 
Find The Limit $$\lim_{x\to0}\left(\frac{\tan(x)}{x}\right)^{\displaystyle\frac{1}{x^{2}}}$$

My Approach
We know that $$\lim_{x\to0}\frac{\tan(x)}{x}
= 1$$
So $$\lim_{x\to0}
\left[ 1+\left(\frac{\tan(x)}{x}-1\right)\right]
^{\displaystyle\left(\frac{1}{\left(\frac{\tan(x)}{x}-1\right)}\right)\left(\frac{\tan(x)}{x}-1\right)\left(\frac{1}{x^{2}}\right)}
=e^{\left[\displaystyle\lim_{x\to0}\left(\frac{\tan(x)}{x}-1\right)\left(\frac{1}{x^{2}}\right)\right]}$$
From here I tried many methods but couldn’t get the solution. I also tried
the logrithmic approach but everything got messesed up.
 A: $$\lim_{x\rightarrow0}\left(\frac{\tan{x}}{x}\right)^{\frac{1}{x^2}}=\lim_{x\rightarrow0}\left(1+\frac{\tan{x}}{x}-1\right)^{\frac{1}{\frac{\tan{x}}{x}-1}\frac{\frac{\tan{x}}{x}-1}{x^2}}=$$
$$=e^{\lim\limits_{x\rightarrow0}\left(\frac{\sin{x}-x\cos{x}}{x^3}\cdot\frac{1}{\cos{x}}\right)}=e^{\lim\limits_{x\rightarrow0}\frac{\cos{x}-\cos{x}+x\sin{x}}{3x^2}}=e^{\frac{1}{3}}.$$
A: Form the known form $\lim_{x\to 0} f (x)^{g (x)}=1^{\infty} $ the limit is given by $e^ {\lim_{x\to 0} (f (x)-1).g (x)} $ so in your case it will be $e^{\lim_{x\to 0} \frac {\tan(x)-x}{x^3}} $ now as given in comments use Taylor series of $\tan (x) $ to get the answer as $e^{\frac {1}{3 }} $
A: $$\lim_{x\to0}\left(\frac{\tan(x)}{x}\right)^{\frac{1}{x^{2}}}=e^{\lim_{x\to0}\ln\left(\left(\frac{\tan(x)}{x}\right)^{\frac{1}{x^{2}}}\right)}=e^{\lim_{x\to0}\frac{\ln\left(\frac{\tan(x)}{x}\right)}{x^2}}=e^{\frac{0}{0}}\\\lim_{x\to0}\frac{\ln\left(\frac{\tan(x)}{x}\right)}{x^2}=\lim_{x\to0}-\dfrac{\tan\left(x\right)-x\sec^2\left(x\right)}{2x^2\tan\left(x\right)}=-\lim_{x\to0}\frac{-2x\sec^2\left(x\right)\tan\left(x\right)}{4x\tan\left(x\right)+2x^2\sec^2\left(x\right)}=\\-\lim_{x\to0}\frac{-4x\sec^2\left(x\right)\tan^2\left(x\right)-2\sec^2\left(x\right)\tan\left(x\right)-2x\sec^4\left(x\right)}{4x^2\sec^2\left(x\right)\tan\left(x\right)+4\tan\left(x\right)+8x\sec^2\left(x\right)}=\\-\lim_{x\to0}\frac{-8x\sec^2\left(x\right)\tan^3\left(x\right)-8\sec^2\left(x\right)\tan^2\left(x\right)-16x\sec^4\left(x\right)\tan\left(x\right)-4\sec^4\left(x\right)}{8x^2\sec^2\left(x\right)\tan^2\left(x\right)+24x\sec^2\left(x\right)\tan\left(x\right)+4x^2\sec^4\left(x\right)+12\sec^2\left(x\right)}\\=-\frac{-4}{12}=\frac{1}{3}\\\therefore\lim_{x\to0}\left(\frac{\tan(x)}{x}\right)^{\frac{1}{x^{2}}}=e^{\frac{1}{3}}$$
