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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?

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

2 Answers 2

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$$\lim_{x \to 0}\frac{\sin(6x)}{8x}=\lim_{x \to 0}\frac{\sin(6x)}{6x}\frac{6}{8}=\frac{3}{4}.$$

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Hint: Use Taylor series: $\sin(6x)=6x+O(x^{3})$ as $x \rightarrow 0$.

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  • $\begingroup$ I haven't learnt Taylor series yet. Could you do it in full steps. Thank you $\endgroup$ Feb 19, 2019 at 1:53
  • $\begingroup$ @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. $\endgroup$
    – JMoravitz
    Feb 19, 2019 at 1:55
  • $\begingroup$ @JamesWarthington: I misunderstood why you were having trouble. You can ignore my answer if you already know that $\sin(x)/x \rightarrow 1$ as $x\rightarrow 0$. $\endgroup$
    – parsiad
    Feb 19, 2019 at 2:09
  • $\begingroup$ @JMoravitz: Can you do this by expanding $sin(6x)$ into its Taylor series. I am opened to more advanced way of doing this excercise. $\endgroup$ Feb 19, 2019 at 2:11
  • $\begingroup$ @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. $\endgroup$
    – JMoravitz
    Feb 19, 2019 at 2:15

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