Calculate $\lim\limits_{x\to 0}\left(\frac{\sin(x)}x\right)^{1/x^2}$ 
How to calculate 
  $$\lim_{x\to 0}\left(\frac{\sin(x)}x\right)^{1/x^2}?$$

I know the result is $1/(6e)$.
 A: L'Hospital's Rule can be used here, after using the natural log.
\begin{align}
\lim\limits_{x\to0}\left(\frac{\sin(x)}{x}\right)^{1/x^2}& = \lim_{x\to0} \exp(\dfrac{\ln(\sin(x)/x)}{x^2})\\
& = \lim_{x\to0} \exp(\dfrac{\ln(\sin(x))-\ln(x)}{x^2})\\
\lim_{x\to0}\dfrac{\ln(\sin(x))-\ln(x)}{x^2}& = \lim_{x\to0}\dfrac{\frac{\cos(x)}{\sin(x)}-\frac1x}{2x}\\
& = \lim_{x\to0}\dfrac{\frac{x\cos(x)-\sin(x)}{x\sin(x)}}{2x}\\
& = \lim_{x\to0}\dfrac{x\cos(x)-\sin(x)}{2x^2\sin(x)}\\
&= \lim_{x\to0}\dfrac{-x\sin(x)}{4x\sin(x)+2x^2\cos(x)}\\
&= \lim_{x\to0}\dfrac{-\sin(x)}{4\sin(x)+2x\cos(x)}\\
&=\lim_{x\to0}\dfrac{-\cos(x)}{6\cos(x)-2x\sin(x)}=-\frac16\\
\end{align}
So 
$$\lim\limits_{x\to0}\left(\frac{\sin(x)}{x}\right)^{1/x^2}=\exp(-1/6)=e^{-1/6}$$
A: HINT:
$$\sin x=x-\dfrac{x^3}{3!}+O(x^5)$$
$$\left(\dfrac{\sin x}x\right)^{1/x^2}=\left[\left(1+O(x^4)-\dfrac{x^2}6\right)^{\dfrac1{1+O(x^4)-\dfrac{x^2}6}}\right]^{\dfrac{1+O(x^4)-\dfrac{x^2}6}{x^2}}$$
A: Successive application of L'Hospital's Rule reveals 
$$\begin{align}
\lim_{x\to 0}\frac{\log \left(\frac{\sin x}{x}\right)}{x^2}&=\lim_{x\to 0}\left(\frac{x\cos x-\sin x}{2x^3}\right)\lim_{x\to 0}\left(\frac{x}{\sin x}\right)\\\\
&=\lim_{x\to 0}\frac{-\sin x }{6x}\\\\
&=-\frac16
\end{align}$$
from which we have 

$$\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^{1/x^2}=e^{-1/6}$$

A: In the same spirit as other answers but trying to get some more information $$A=\left(\frac{\sin(x)}x\right)^{\frac 1{x^2}}$$ $$\log(A)=\frac 1{x^2}\,\log\left(\frac{\sin(x)}x\right)$$ Now, using Taylor $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ $$\frac{\sin(x)}x=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^5\right)$$ Now, using $$\log(1-y)=-y-\frac{y^2}{2}-\frac{y^3}{3}+O\left(y^4\right)$$ replace  $y$ by $(\frac{x^2}{6}-\frac{x^4}{120}+\cdots)$, expand just keeping the low powers of $x$; this will give $$\log\left(\frac{\sin(x)}x\right)=-\frac{x^2}{6}-\frac{x^4}{180}+\cdots$$ $$\log(A)=-\frac{1}{6}-\frac{x^2}{180}+\cdots$$ So, $$A=e^{-\frac{1}{6}-\frac{x^2}{180}+\cdots}=e^{-\frac{1}{6}}e^{-\frac{x^2}{180}}=e^{-\frac{1}{6}}(1-\frac{x^2}{180})$$ which shos the limit and how it is reached.
A: Re-write the logarithm of the original expression as follows
\begin{eqnarray*}
\ln \left( \frac{\sin x}{x}\right) ^{1/x^{2}} &=&\frac{1}{x^{2}}\ln \left(
1+\left( \frac{\sin x}{x}-1\right) \right)  \\
&=&\frac{\ln \left( 1+\left( \frac{\sin x}{x}-1\right) \right) }{\left( 
\frac{\sin x}{x}-1\right) }\frac{\left( \frac{\sin x}{x}-1\right) }{x^{2}} \\
&=&\frac{\ln \left( 1+\left( \frac{\sin x}{x}-1\right) \right) }{\left( 
\frac{\sin x}{x}-1\right) }\frac{\left( \sin x-x\right) }{x^{3}}
\end{eqnarray*}
Now using the standard limits
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x}{x}=1,\ \ \ \ \ \ \ \ \ \lim_{u\rightarrow
0}\frac{\ln (1+u)}{u}=1,\ \ and\ \ \ \lim_{x\rightarrow 0}\frac{\sin x-x}{%
x^{3}}=-\frac{1}{6}.
\end{equation*}
It follows that
\begin{equation*}
\lim_{x\rightarrow 0}\ln \left( \frac{\sin x}{x}\right) ^{1/x^{2}}=1\cdot
\left( -\frac{1}{6}\right) =-\frac{1}{6}.
\end{equation*}
Taking back the exponential one gets
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{\sin x}{x}\right) ^{1/x^{2}}=e^{-1/6}.
\end{equation*}
A: Let us expand $sin(x)$ for small values of $x$. It is sufficient to take the first two terms, so we get $x - x^3/6$. Dividing by $x$ yields $1 - x^2/6$ for the term between brackets. Taking this to the power $1/x^2$, and applying the well-known formula for the exponential function, yields the answer: $e^{-1/6}$.  
