<s>[INST] 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.
 [/INST] $\require{cancel}$
$$\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$$
$$f(x):=\left(\frac{\sin x}x\right)^{1/{x^2}}$$
$$\ln f(x)= \dfrac{1}{x^2}\ln \left( \dfrac{\sin x}{x}-1+1\right)$$
$$\ln f(x)=\dfrac{\ln \left( \dfrac{\sin x}{x}-1+1\right)}{\dfrac{\sin x}{x}-1}\times \dfrac{\dfrac{\sin x}{x}-1}{x^2}$$
$$\ln f(x)=\dfrac{\ln \left( \dfrac{\sin x}{x}-1+1\right)}{\dfrac{\sin x}{x}-1}\times \dfrac{\sin x-x}{x^3}$$
$$\lim_{x \to 0} \ln f(x)=\lim_{x\to 0}\cancelto{1}{\dfrac{\ln \left( \dfrac{\sin x}{x}-1+1\right)}{\dfrac{\sin x}{x}-1}}\times \cancelto{\dfrac{-1}{6}}{\dfrac{\sin x-x}{x^3}}=\dfrac{-1}{6}$$
$$\lim_{x \to 0} f(x)=\text{exp}(-\dfrac{1}{6})$$

Used from
Calculate the limit : $\lim_{x \to 0}\frac{x-\sin{x}}{x^3}$ WITHOUT using L'Hopital's rule
Limit of a function without using L'Hôpital Rule
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