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.

THIS QUESTION'S "BEST" ANSWER, uses basic tools only to establish the inequalities for the sine function
$$\cos z\le \frac{\sin z}{z}\le 1 \tag 1$$
for $-\pi/2 \le z\le \pi/2$.  
Then, letting $z=2x$ in $(1)$ reveals
$$\cos (2x)\le \frac{\sin(2x)}{2x}\le 1 \tag 2$$
for $-\pi/4\le x\le \pi/4$.  
Dividing both sides of $(2)$ by $4$ we obtain
$$\frac14 \cos (2x)\le \frac{\sin(2x)}{8x}\le \frac14 $$
whereupon applying the Squeeze Theorem yields
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\frac{\sin(2x)}{8x}=\frac14}$$
as expected!
