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 ?

Solution 1:

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 $e^{-\frac16}$.

Solution 2:

$$\ln(y)=\lim_{x \rightarrow 0} \frac{1}{x^2}\ln\left(\frac{\sin(x)}{x}\right)$$
Becomes: $$\frac{x\cos(x)-\sin(x)}{2x^2\sin(x)}$$ After applying L'Hôpital's rule again differentiating two more times w.r.t $x$, we get:
$$\lim_{x \rightarrow 0} \frac{-\cos(x)}{-2x\sin(x)+2\cos(x)+4\cos(x)}=-1/6$$