Finding the limit $\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$ [duplicate]

Working with $\exp\left(\frac1{x^2}\ln\frac{\sin x}x\right)$ and l'Hopital should be fine. Since $\frac{\sin x}x\to 1$ we have a "$\frac 00$" case here:

$$\lim_{x\to 0}\frac{\ln\frac{\sin x}x}{x^2}=\lim_{x\to0}\frac{\frac x{\sin x}\frac d{dx}\frac{\sin x}x}{2x}=\lim_{x\to0}\frac{{x\cos x-\sin x}}{2x^2\sin x}$$ Apply l'Hopital again and cancel one $x$ $$\begin{align}\lim_{x\to0}\frac{{x\cos x-\sin x}}{2x^2\sin x}&=\lim_{x\to0}\frac{-x\sin x}{4x\sin x+2x^2\cos x}\\ &=\lim_{x\to0}\frac{-\sin x}{4\sin x+2x\cos x}\end{align}$$ and l'Hopital again $$\lim_{x\to0}\frac{-\sin x}{4\sin x+2x\cos x} =\lim_{x\to0}\frac{-\cos x}{4\cos x+2\cos x-2x\sin x}=-\frac16. $$ By applying the $\exp$ again, we obtain $$ \lim_{x\to 0}\left(\frac1{x^2}\ln\frac{\sin x}x\right)^{\frac1{x^2}}=e^{-1/6}.$$

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 $x\to0$

$\displaystyle\implies\lim_{x\to0}\frac{x^2}{-\frac{x^2}{3!}+O(x^4)}=-\frac16$

Here is a method beased on Taylor series. Using the Taylor series of $\frac{\sin x}{x}$, we get

$$ e^{\frac{1}{x^2}\ln(1-x^2/3!+\dots) }= e^{\frac{1}{x^2}\ln(1-t) },$$

where $t=\frac{x^2}{3!}-\frac{x^4}{5!}+\dots$. Using the Taylor series of $\ln(1-t)$, we have

$$e^{\frac{1}{x^2}\ln(1-t) }=e^{\frac{1}{x^2}(-t-t^2/2-\dots) } = e^{\frac{1}{x^2}(-(x^2/3!-x^4/5!+\dots)-(x^2/3!-x^4/5!+\dots)^2/2-\dots) }\longrightarrow_{x\to 0} e^{-\frac{1}{3!}} $$

Note: We used the following Taylor series

$$ \frac{\sin x}{x}=1-\frac{x^2}{3!}+\dots, $$

$$ \ln(1-t)=-t-\frac{t^2}{2}- \frac{t^3}{3}-\dots\,. $$