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?
 A: 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
  $$ \lim_{x\to 0+} \left(\frac{\sin x }{x} \right)^\frac{1}{x}= \lim_{x\to 0+} \exp\left( \frac{1}{x} \ln \left(\frac{\sin x }{x} \right)\right) = \exp \left( \lim_{x\to 0+} \ln \left(\frac{\sin x }{x}\right) \right) = \exp 0 =1.$$

A: Using Taylor polynomials, 
$$
\frac1x\,\log\frac {\sin x}x=\frac1x\,\log (1-O (x^2))=O (x)\to0.
$$
A: Using the big-O notation in its common sense,
$$\sin x = x + O(x^3),$$
whence
$$\ln(\frac {\sin x}x)=\ln\frac{x+O(x^3)}x=\ln(1+O(x^2))=O(x^2).$$
A: Since $f(x)=e^x$ is a continuous function, we obtain:
$$\lim_{x\to 0^+}\left(\frac{\sin{x}}{x}\right)^{\frac{1}{x}}=\lim_{x\to 0^+}\left(1+\frac{\sin{x}}{x}-1\right)^{\frac{1}{\frac{\sin{x}}{x}-1}\frac{\frac{\sin{x}}{x}-1}{x}}=e^{\lim\limits_{x\rightarrow0}\frac{\sin{x}-x}{x^2}}=1$$
Because $\lim\limits_{x\rightarrow0}\frac{\sin{x}-x}{x^2}=\lim\limits_{x\rightarrow0}\frac{\cos{x}-1}{2x}=0$
A: There is an interesting massive overkill: we may prove that in a right neighbourhood of the origin we have
$$1-\frac{x^2}{6}\leq\frac{\sin x}{x}\leq \exp\left(-\frac{x^2}{6}\right) \tag{1}$$
hence the wanted limit is trivially $\exp(0)=1$. By the Weierstrass product for the sine function:
$$ \log\left(\frac{\sin x}{x}\right) = \sum_{n\geq 1}\log\left(1-\frac{x^2}{\pi^2 n^2}\right)\leq\sum_{n\geq 1}-\frac{x^2}{\pi^2 n^2}=-\frac{\zeta(2)}{\pi^2}x^2=-\frac{x^2}{6}\tag{2} $$
and we are done, since the inequality on the left of $(1)$ just follows from the Taylor series of $\text{sinc}(x)$.
