<s>[INST] How do I solve this limit: $\lim _{x \to 0} \left(\frac{ \sin x}{x}\right)^{1/x}$? $$\lim _{x \rightarrow 0} \left(\frac{
\sin x}{x}\right)^{1/x}$$
I have spent an hour on the above limit and have no work to show. I tried using L'Hopital's Rule, but just kept going around in circles. Any help would be appreciated. Thank you.
 [/INST] Hint : Take its logarithm, use the fact that $\ln a^b=b\ln a=\dfrac{\ln a}{1/b}$, and apply l'Hopital $3$ or $4$ times.

$$\ln L=\lim_{x\to0}\frac1x\cdot\ln\frac{\sin x}x=\lim_{x\to0}\frac{\ln\sin x-\ln x}x=\lim_{x\to0}\frac{\dfrac{\cos x}{\sin x}-\dfrac1x}1=\lim_{x\to0}\frac{x\cdot\cos x-\sin x}{x\cdot\sin x}=$$
$$=\lim_{x\to0}\frac{(1\cdot\cos x-x\cdot\sin x)-\cos x}{1\cdot\sin x+x\cdot\cos x}=-\lim_{x\to0}\frac{x\cdot\sin x}{\sin x+x\cdot\cos x}=$$
$$=-\lim_{x\to0}\frac{1\cdot\sin x+x\cdot\cos x}{\cos x+(1\cdot\cos x-x\cdot\sin x)}=-\lim_{x\to0}\frac{\sin x+x\cdot\cos x}{2\cdot\cos x-x\cdot\sin x}=-\frac{0+0\cdot0}{2\cdot1-0\cdot0}=-\frac02$$
$$=0\iff L=e^0=1.$$
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