<s>[INST] 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 $1/(6e)$.
 [/INST] 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.
</s>