Finding the limit $\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$ 
Find the following limit: $$\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$$

Well I tried to do the $\exp\left(\frac{ \ln\frac{\sin x}{x}}{x^2}\right)$ then apply LHR but I seem to get to endless dervivations... 
There's got to be a more simple approach.

Using Taylor Series ,
$$\sin x=x-\frac{x^3}{3!}+O(x^5)\implies \frac{\sin x}x=1-\frac{x^2}{3!}+O(x^4)$$
$$\implies \lim_{x\to0}\left(\frac{\sin x}x\right)^{\frac1{x^2}}=\lim_{x\to0} \left(1-\frac{x^2}{3!}+O(x^4)\right)^{\frac1{x^2}}$$
$$=\left(\lim_{x\to0}\left(1-\frac{x^2}{3!}+O(x^4)\right)^{\frac1{-\frac{x^2}{3!}+O(x^4)}}\right)^{\lim_{x\to0}\frac{-\frac{x^2}{3!}+O(x^4)}{x^2}}$$
Now if we set $\displaystyle -\frac{x^2}{3!}+O(x^4)=-\frac1u,$ the inner limit reduces to $\displaystyle\lim_{u\to\infty}\left(1+\frac1u\right)^u=e$
For the exponent,  $\displaystyle\lim_{x\to0}\frac{-\frac{x^2}{3!}+O(x^4)}{x^2}=\lim_{x\to0}\left({-\frac1{3!}+O(x^2)}\right)$ as $x\ne0$ as $x\to0$
$\displaystyle\implies\lim_{x\to0}\frac{x^2}{-\frac{x^2}{3!}+O(x^4)}=-\frac16$
