# 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 $\lim_{x \to 0} \frac{\sin(2x)}{8x} = \frac{0}{0}$?

• Use l'hopital's rule once and you will see the limit is $\frac{1}{4}$. Jun 30, 2014 at 1:04
• $\frac{0}{0}$ is not a number. If you get this as an answer, it means that you have not actually got an answer at all. It might be that the problem does not have a sensible answer, or it might be that you need to approach the question by different methods. See the answer below for a possibility. Jun 30, 2014 at 1:18
• They will want something like the answers below, but when $t$ is real close to $0$, $(\sin t)/t$ is almost $1$, so $\sin(2x)/(8x)$ is almost $2x/8x$, that is, almost $1/4$. Jun 30, 2014 at 2:14
• Getting $\frac{0}{0}$ as an answer tells you that the numerator and the denominator will both get smaller and smaller. But they don't get smaller at the same rate, so the answer $\frac{0}{0}$ isn't useful at all. By applying some clever manipulation, you get to the actual correct answer, $\frac{1}{4}$, which tells you that both the numerator and the denominator will get arbitrarily small, but the denominator will always be approximately $4$ times larger than the numerator. Jun 30, 2014 at 3:34
• That is, if you get $\frac{0}{0}$ as an answer, try again using another approach, since you'll probably be able to get more information from the limit. Jun 30, 2014 at 3:35

One very common limit is: $$\lim_{u\to0}\dfrac{\sin u}u=1.$$ You can use it here by noting that: $$\lim_{x \to 0} \frac{\sin(2x)}{8x} =\lim_{x \to 0} \dfrac14\frac{\sin(2x)}{(2x)} = \ldots$$

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:

$$\lim_{\delta \rightarrow 0} \dfrac{\sin 2\delta}{8\delta} = \lim_{\delta \rightarrow 0} \dfrac{2 \cos 2\delta}{8} = \dfrac14$$

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.