Originally Posted by

**redsoxfan325** Sorry, I actually mislead you the way I wrote that. I wrote $\displaystyle \frac{\sin(4x^{-2})}{4x^{-2}}$ thinking that it would fall under the $\displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta}=1$ category, but it doesn't because $\displaystyle \lim_{x\to0}4x^{-2}\neq0$.

So the way you should actually do the limit is to use the squeeze theorem.

$\displaystyle \lim_{x\to0}-x^2e^{-x^2/4}\leq\lim_{x\to0}x^2e^{-x^2/4}\sin(4x^{-2})\leq\lim_{x\to0}x^2e^{-x^2/4}$

Thus, $\displaystyle 0\leq\lim_{x\to0}x^2e^{-x^2/4}\sin(4x^{-2})\leq0$, so the limit is zero.

Sorry for any confusion I may have caused.