Evaluating $ \lim_{x \rightarrow 0}\left (\frac{\sin(x)}{x}\right)^\frac{1}{x^2}$ I have got this problem on limits 
$$ \lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x}\right)^\frac{1}{x^2}$$
What I am doing here is that taking log and then applying L-Hospital's Rule
$$y= \lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x}\right)^\frac{1}{x^2}
$$
then $$\ln(y)=\lim_{x \rightarrow 0} \frac{1}{x^2}\ln\left(\frac{\sin(x)}{x}\right)$$ now it becomes $\frac{0}{0}$ type , but problem is how to take derivative here? when differentiating $\ln\left(\frac{\sin(x)}{x}\right)$ it will become $\frac{x}{\sin(x)}$ then we'll have to differentiate one more time like we do in case of $f(f(x))$ ? $\sin(x)$ and $x$ will be separately differentiated or together ?
Use the Taylor series to get
$$\lim_{x\to0}\frac1{x^2}\ln\left(\frac{\sin x}{x}\right)=\lim_{x\to0}\frac1{x^2}\ln\left(\frac{ x-x^3/6+o(x^3)}{x}\right)=\lim_{x\to0}\frac1{x^2}\ln(1-x^2/6+o(x^2))\\=\lim_{x\to0}\frac1{x^2}(-x^2/6+o(x^2))=-\frac16$$
hence the desired limit is $e^{-\frac16}$.
Using $$\displaystyle \sin (\pi z) = \pi z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right)\;,$$ Then Put $\displaystyle \pi z= x\Rightarrow z=\frac{x}{\pi}$

So $$\displaystyle \lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^2}} = \lim_{x\rightarrow 0}\prod_{n=1}^{\infty}\left(1-\frac{x^2}{\pi^2n^2}\right)^{\frac{1}{x^2}}$$

Now we can Interchange Product and Limit, We get

$$\displaystyle \lim_{x\rightarrow 0}\left(\frac{\sin x}{x}\right)^{\frac{1}{x^2}} = \prod_{n=1}^{\infty}\lim_{x\rightarrow 0}\left(1-\frac{x^2}{\pi^2n^2}\right)^{\frac{1}{x^2}} = \prod_{n=1}^{\infty}e^{-\frac{1}{\pi^2n^2}} = e^{-\sum_{n=1}^{\infty}\frac{1}{\pi^2n^2}} = e^{-\frac{1}{6}}$$