Finding the limit of $\frac {\sin(2x)} {8x}$ I am taking an online course in Calculus from Ohio State and am just being introduced to the concept of limits. One of the exercises given to me is to find the limit of the following
$$\lim_{x \rightarrow 0}\frac {\sin({2x})}{8x} $$
Using what I have so far been taught, I determined that this is equivalent to the following
$$\frac {\lim_{x \rightarrow 0}\sin(2x)} {(\lim_{x \rightarrow 0}8) (\lim_{x \rightarrow 0}x)} $$
As far as I am aware, this should become
$$\frac 0 {(8)(0)}$$
which is undefined, so I have a feeling this is incorrect. I'm sure that this is an easy problem to solve, I'm just not sure how to do it without getting $\frac 0 0$ as an answer.

Here's a case where you use L'Hôpital's rule.
Given two functions $f(x)$ and $g(x)$, if
$$\lim_{x\to c}\frac{f(x)}{g(x)}$$
is indeterminate (for all intents and purposes, undefined in this case), then
$$\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$$
and the same for higher derivatives.
Therefore, in this case, all you have to do is take the derivative of the top and the derivative of the bottom, and find the limit as $x\to0$.
$$\lim_{x\to0}\frac{\sin(2x)}{8x}=\lim_{x\to0}\frac{\frac{\mathrm{d}}{\mathrm{d}x}(\sin(2x))}{\frac{\mathrm{d}}{\mathrm{d}x}(8x)}$$
Can you simplify from here?
