Why is $\lim_{x \to 0} \frac{\sin(2x)}{8x} = \frac{1}{4}$ If $\lim_{x \to 0} \sin(2x) = 0$ and $\lim_{x \to 0} 8x = 0$, then isn't $\lim_{x \to 0} \frac{\sin(2x)}{8x} = \frac{0}{0}$?
 A: Given that $f(0) = g(0) = 0$.
$$ \begin{align}
\\ \lim_{\delta \rightarrow 0} \dfrac{f(\delta)}{g(\delta)} & = \lim_{\delta \rightarrow 0} \dfrac{f(\delta) - f(0)}{g(\delta) - g(0)}
\\ & = \lim_{\delta \rightarrow 0} \dfrac{\dfrac{f(0 + \delta) - f(0)}{\delta}}{\dfrac{g(0 + \delta) - g(0)}{\delta}}
\\ & = \lim_{\gamma \rightarrow 0} \left( \lim_{\delta \rightarrow 0} \dfrac{\dfrac{f(\gamma + \delta) - f(\gamma)}{\delta}}{\dfrac{g(\gamma + \delta) - g(\gamma)}{\delta}}\right)
\\ &= \lim_{\gamma \rightarrow 0} \dfrac{f'(\gamma)}{g'(\gamma)}
\end{align}$$
This is called L'Hopital's rule.
So when you have an apparently indeterminate limit form that looks like $\dfrac{0}{0}$, you can take the derivative of the numerator and denominator without changing the limit.
So:
$$\lim_{\delta \rightarrow 0} \dfrac{\sin 2\delta}{8\delta} = \lim_{\delta \rightarrow 0} \dfrac{2 \cos 2\delta}{8} = \dfrac14$$
A: You should have shown in your Calculus class that 
$$\lim\limits_{x \to 0}\dfrac{\sin(x)}{x} = 1\text{.}$$
You can't change the part that is inside the $\sin$ function, so you might want to use the fact that 
$$\lim\limits_{2x \to 0}\dfrac{\sin(2x)}{2x} = 1\text{.}$$
But if $2x \to 0$, then $x \to 0$. So in other words, what you now know is that 
$$\lim\limits_{x \to 0}\dfrac{\sin(2x)}{2x} = 1\text{.}$$
To get what you want, multiply both sides of this equality by $\dfrac{1}{4}$. 
$$\begin{align}\dfrac{1}{4}\lim\limits_{x \to 0}\dfrac{\sin(2x)}{2x} = \dfrac{1}{4} \Leftrightarrow \lim\limits_{x \to 0}\dfrac{\sin(2x)}{8x} = \dfrac{1}{4}\text{.}\end{align}$$
From my experience in being a TA for a Calculus I course, I prefer this more general method since it can be used to find seemingly different limits (to someone new to the material), such as
$$\lim\limits_{x \to 3}\dfrac{\sin(x-3)}{x-3}$$
as a quick example off the top of my head.
A: One very common limit is: $$\lim_{u\to0}\dfrac{\sin u}u=1.$$
You can use it here by noting that: $$\lim_{x \to 0} \frac{\sin(2x)}{8x} =\lim_{x \to 0} \dfrac14\frac{\sin(2x)}{(2x)} = \ldots$$
