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.
 A: 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.

$$\ln L=\lim_{x\to0}\frac1x\cdot\ln\frac{\sin x}x=\lim_{x\to0}\frac{\ln\sin x-\ln x}x=\lim_{x\to0}\frac{\dfrac{\cos x}{\sin x}-\dfrac1x}1=\lim_{x\to0}\frac{x\cdot\cos x-\sin x}{x\cdot\sin x}=$$
$$=\lim_{x\to0}\frac{(1\cdot\cos x-x\cdot\sin x)-\cos x}{1\cdot\sin x+x\cdot\cos x}=-\lim_{x\to0}\frac{x\cdot\sin x}{\sin x+x\cdot\cos x}=$$
$$=-\lim_{x\to0}\frac{1\cdot\sin x+x\cdot\cos x}{\cos x+(1\cdot\cos x-x\cdot\sin x)}=-\lim_{x\to0}\frac{\sin x+x\cdot\cos x}{2\cdot\cos x-x\cdot\sin x}=-\frac{0+0\cdot0}{2\cdot1-0\cdot0}=-\frac02$$
$$=0\iff L=e^0=1.$$
A: Let $$
 y = \frac{\sin x}{x}-1
$$ 
and notice that for $x\to 0$ also $y\to 0$. Then
remember that
$$
\lim_{y\to 0} \left(1+y\right)^{\frac 1 y} = e
$$
while (use Hopital or Taylor here)
$$
\frac{y}{x} = \frac{\frac{\sin x}{x} - 1}{x} = \frac{\sin x - x}{x^2} \to 0.
$$
So your limit is
$$
 (1+y)^\frac 1x = \left((1+y)^\frac{1}{y}\right)^{\frac{y}{x}} \to e^0=1
$$
A: For small values of $x$, the Taylor series for $\sin(x)$ is $x - x^3 / 6$; then, for $\sin(x)/ x$, the expansion around $x=0$ is $1 - x^2 /6$. Now, remember that, for small values of $y$, $(1+y)^a$ is approximated by $1 + a y$ (another Taylor series). So, for your expression, you arrive to 

$$(1 - x^2 /6)^{1/x}$$ 

which is almost $(1 - x / 6)$. So, your limit is $1$.
Edit
Probably better (at least to me three years after my first answer), considering
$$ A=\left(\frac{
\sin x}{x}\right)^{1/x}\implies \log(A)=\frac 1x \log\left(\frac{
\sin x}{x}\right)$$ Now, using Taylor series around $x=0$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^5\right)$$ $$\frac{
\sin x}{x}=1-\frac{x^2}{6}+O\left(x^4\right)$$ $$\log\left(\frac{
\sin x}{x}\right)=-\frac{x^2}{6}+O\left(x^4\right)$$ $$\log(A)=\frac 1x \log\left(\frac{
\sin x}{x}\right)=-\frac{x}{6}+O\left(x^3\right)$$ Now, using $A=e^{\log(A)}$ and Taylor again $$A=1-\frac{x}{6}+\frac{x^2}{72}+O\left(x^3\right)$$ Unsing $x=\frac \pi 6$, the "exact" value $$\left(\frac{3}{\pi }\right)^{6/\pi }\approx 0.915689$$ while the above asymptotics would give $$1-\frac{\pi }{36}+\frac{\pi ^2}{2592}\approx 0.916541$$
A: Notice by Young's Inequality since $(\frac{1}{x} + (1 - \frac{1}{x})) = 1$, then
$$ (\frac{\sin x}{x} )^{1/x}=(\frac{\sin x}{x} )^{1/x} 1^{1 - \frac{1}{x}} \leq \frac{\sin x}{x^2} + 1 - \frac{1}{x} = \frac{\sin x - x}{x^2} + 1$$
Now, for positive $x$ and for $a \leq 1$, we have that $a^x \leq x +1 $. Hence $a^{1/x} \geq \frac{1}{x+1}$. Now, since $\frac{\sin x}{x} \leq 1$, we apply this inequality with $a = \frac{\sin x}{x} $ to obtain
$$  (\frac{\sin x}{x} )^{1/x} \geq \frac{1}{x+1}$$. Hence we have
$$  \frac{1}{x+1} \leq (\frac{\sin x}{x} )^{1/x} \leq \frac{\sin x - x}{x^2} + 1$$.
Now, since $\lim{ \frac{1}{x+1} } = 1 $ and 
$$ \lim (\frac{\sin x - x}{x^2} + 1 ) = \lim ( \frac{\sin x - x}{x^2} ) = 1 + \lim ( \frac{\cos x - 1}{2x} ) = 1 + \lim ( \frac{- \sin x}{2} ) = 1 $$
Now, result follows by the squeeze trick.
A: $\lim\limits_{x\to0}(\frac{\sin{x}}{x})^{\frac{1}{x}}=\lim\limits_{x\to0}e^{\frac{\ln{\frac{\sin{x}}{x}}}{x}}=$ $e^{\lim\limits_{x\to0}{\frac{\ln{\frac{\sin{x}}{x}}}{x}}}=\frac{0}{0}$ $=e^{\lim\limits_{x\to0}\frac{x}{\sin{x}}\frac{x\cos{x}-\sin{x}}{x^2}}$, so you were right, it doesn't simplify nicely. I thought it would, sorry. If you are familiar with ~ and o, you can finish this using $\sin{x}$~$x-\frac{x^3}{6},x\to0$.
A: A suggestion to showing work for this problem would be to show that lim x>0 of sinx/x=1 and then just simply applying the definition of the limit.(you will end up with 1^(1/x) and since 1 is unitary 1 to any power is just 1.
A: For another version of the Taylor series approach, note that (taking the first couple of terms of the Taylor series for $\sin x$ and dividing by $x$) we come to the same limit $(1-x^2/6)^{1/x}$ that Claude Leibovici gets; setting $y=\frac1x$ we get the limit as $\displaystyle\lim_{y\to\infty}(1-\frac1{6y^2})^y$ $\displaystyle=\lim_{y\to\infty}(1-\frac1{6y^2})^{(6y^2)\cdot1/(6y)}$ $\displaystyle=\lim\left[(1-\frac1{6y^2})^{6y^2}\right]^{1/(6y)}$.  Now, the limit in brackets goes to a finite value ($e^{-1}$) as $y\to\infty$, and the exponent outside the brackets goes to zero, so the overall limit is $(e^{-1})^0=1$.
