find the limit of $\lim_{x \to 0} \frac{1}{8}\sin(x)^{1/x^2}$ 
$$\lim_{x \to 0} \frac{\sin(x)}{8}^{\frac{1}{x^2}}$$

$y=\lim_{x \to 0}\frac{1}{x^2}\cdot \lim_{x \to 0} \ln(\frac{\sin(x)}{8})=-\infty$
So we have $e^{-\infty}=0$?
 A: The limit is $0$.
We have for $0 < x < 1$ the inequality
$$-\log x = \int_x^1 \frac{dt}{t} \geqslant \frac{1-x}{1} \\ \implies \log x \leqslant x - 1$$
If $x > 0$, then as $x \to 0+$
$$\frac{\log \sin x}{x^2} \leqslant \frac{\sin x - 1}{x^2} \to -\infty.$$
If $x < 0$, then, extending into the complex plane,  as $x \to 0-$
$$\frac{\log \sin x}{x^2} = \frac{\log\left( -\sin |x|\right)}{|x|^2} = \frac{\log\left( \sin |x|\right) + \log(e^{i\pi})}{|x|^2} \to -\infty.$$
Hence
$$\lim_{x \to 0} (\sin x)^{1/x^2} = \exp\left(\lim_{x \to 0}\frac{\log \sin x}{x^2} \right) = \exp(-\infty) = 0$$
A: No the limit does not exist because you don't know sin(x) is positive or negative around 0
A: The function is not defined for $x<0$ because of the problem of raising negative numbers to arbitrary powers. For $x>0,$ we have $0<\sin x < 1/2$ for $0< x< \pi/6.$ So for small $x>0,$ 
$$0<(\sin x)^{1/x^2} < (1/2)^{1/x^2}.$$
Because $0<1/2<1$ and $1/x^2 \to \infty,$ $ (1/2)^{1/x^2} \to 0$ as $x\to 0^+.$ Thus our limit (modified to require $x\to 0^+$) is $0$ by the squeeze theorem. 
