<s>[INST] Evaluate $\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$ I need to find the following limit: $$\lim_{x\to 0^+}\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}$$
I started this way: $$\left(\frac{\sin x}{x}\right)^{\frac{1}{x}}=e^{\frac{1}{x}\cdot \ln\left[\frac {\sin x}{x}\right]}$$
So it's enough to find: $$\lim_{x\to 0^+}\frac{\ln(\frac {\sin x}{x})}{x}$$
I tried to use L'Hôpital's rule but it got me nowhere. Any ideas?
 [/INST] Hints: 


*

*Write
$$\left(\frac{\sin x }{x} \right)^\frac{1}{x} = \exp\left( \frac{1}{x} \ln \left(\frac{\sin x }{x} \right)\right).$$

*$\exp$ is continuous, $\lim_{x\to 0} \frac{\sin x }{x}=1$ and $\ln 1=0$.

*Evaluate $\lim_{x\to 0}\frac{1}{x} \ln \left(\frac{\sin x }{x} \right)$ by using L'Hospital.


For step 3 use $$\ln \left(\frac{\sin x }{x} \right)' =\frac{\cos x}{\sin x} - \frac{1}{x} = \frac{x\cos x - \sin x}{x \sin x}.$$
For $x\to 0+$ we have an expression like $0/0$ again. Now apply L'Hospital again:
$$\lim_{x\to 0+}\frac{x\cos x-\sin x}{x\sin x}=\lim_{x\to 0+}\frac{(\cos x-x\sin x)-\cos x}{\sin x+x\cos x}=-\lim_{x\to 0+}\frac{x\sin x}{\sin x+x\cos x}$$
$$=-\lim_{x\to 0+}\frac{\sin x+x\cos x}{\cos x+(\cos x-x\sin x)}=-\lim_{x\to 0+}\frac{\sin x+x\cos x}{2\cos x-x\sin x}=0.$$

So we get
  $$ \lim_{x\to 0+} \left(\frac{\sin x }{x} \right)^\frac{1}{x}= \lim_{x\to 0+} \exp\left( \frac{1}{x} \ln \left(\frac{\sin x }{x} \right)\right) = \exp \left( \lim_{x\to 0+} \ln \left(\frac{\sin x }{x}\right) \right) = \exp 0 =1.$$

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