# How to find the limit $\lim_{x \to 0}\frac{\sin(6x)}{8x}$? [closed]

How do you find the limit $$\lim\limits_{x \to 0}\dfrac{\sin 6x}{8x}$$ ?

I know that I should try to manipulate this expression so it can have the form $$\lim\limits_{x \to 0}\dfrac{\sin6x}{6x}$$ but I don't know how to.

Can you show me the way?

• $\frac{\sin(6x)}{8x} = \frac{\sin(6x)}{6x}\cdot \frac{3}{4}$ Feb 19, 2019 at 1:52

$$\lim_{x \to 0}\frac{\sin(6x)}{8x}=\lim_{x \to 0}\frac{\sin(6x)}{6x}\frac{6}{8}=\frac{3}{4}.$$
Hint: Use Taylor series: $$\sin(6x)=6x+O(x^{3})$$ as $$x \rightarrow 0$$.
• @JamesWarthington all this is is a more rigorous way of reminding you (and the reason why) that $\lim\limits_{x\to 0} \dfrac{\sin(6x)}{6x} = 1$, something which I trust you should already know. Now, just get away from $8$ as the coefficient in the denominator to having $6$ as the coefficient in the denominator using all of the other hints provided. Feb 19, 2019 at 1:55
• @JamesWarthington: I misunderstood why you were having trouble. You can ignore my answer if you already know that $\sin(x)/x \rightarrow 1$ as $x\rightarrow 0$. Feb 19, 2019 at 2:09
• @JMoravitz: Can you do this by expanding $sin(6x)$ into its Taylor series. I am opened to more advanced way of doing this excercise. Feb 19, 2019 at 2:11
• @JamesWarthington The taylor expansion of $\sin$ about zero (or the /definition/ of $\sin$ in many cases if doing so from an analytical point of view) is $\sin(z) = z - \frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\cdots = \sum\limits_{n=0}^\infty \frac{z^{2n+1}}{(2n+1)!}$. Replacing $z$ by $6x$ that gives $\sin(6x) = 6x + O(x^3)$ where $O(x^3)$ is big-oh notation and just means that everything else acts like or is smaller than $x^3$ and "doesn't really matter" in this case. Feb 19, 2019 at 2:15