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I am trying to find the limit of

$$\lim_{x\to 0}\frac{\sin{3x}}{x}$$

I have no idea what I am supposed to do. I know the identity that,

$$\lim_{x\to0}\frac{\sin{x}}{x} = 1$$

but that will not be good enough on a test and I am not sure why that is true anyways. I do not know how I am supposed to proceed with this problem.

J. M. ain't a mathematician
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  • Can you use what you know to figure out $\lim_{n\to\infty}(\sin3x)/(3x)$? – Gerry Myerson May 10 '12 at 13:12
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    @Salech, just so you know: it's a bad idea to have titles that are entirely in $\LaTeX$. Please avoid them. – J. M. ain't a mathematician May 10 '12 at 13:18
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    To editors and suggested edit reviewers: Please do not have titles consist entirely of LaTeX. The operative issue is that then links to this question cannot be "opened in new tab," as the usual right-click window does not appear. – anon May 10 '12 at 13:18
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    Though, in Firefox at least, pressing the shift key and right clicking will pop up the usual context window. – David Mitra May 10 '12 at 13:23
  • @David: still an additional inconvenience, so we prefer that titles not be entirely in $\LaTeX$. – J. M. ain't a mathematician May 10 '12 at 13:26
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    @J.M. I completely agree. But, I thought it might be a good tip for those who weren't aware of it (as I wasn't until recently). – David Mitra May 10 '12 at 13:27
  • Do you know L'Hospital's rule? – MJD May 10 '12 at 13:30
  • Yes but I am not sure if it is suppose to be used at this point in the book. –  May 10 '12 at 13:30
  • You are probably right. – MJD May 10 '12 at 13:32
  • @MarkDominus: I wonder, how do you find the derivative of $\sin$ to use L'Hopital's rule. – SBF May 10 '12 at 14:51
  • @Ilya Booyah – Pedro May 10 '12 at 14:54
  • @PeterTamaroff: I would say that was a trap, but I don't think that the linked proof PLUS the proof of L'Hopital's rule is really easier than the classical proof for $\lim\limits_{x\to 0}\frac{\sin x}{x} = 1$. Moreover, I'm pretty sure that most of the people suggesting L'Hopital's rule for that limit are not paying attention to the fact, how should one find the derivative of $\sin$ – SBF May 10 '12 at 14:59
  • @Ilya I'm not saying it is easier, only that it is possible. =) And you're certainly right on that last remark. – Pedro May 10 '12 at 15:01

4 Answers4

12

Hint: $\dfrac{\sin 3x}{x}=3\dfrac{\sin 3x}{3x}$

Pedro
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Salech Alhasov
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  • I do not see what that does for the problem. –  May 10 '12 at 13:15
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    Replace that $3x$ with $z$, then you'll get: $3\frac{\sin{z}}{z}$, now notice that when $x\to 0$, $z$ tends to zero as well. $3\lim_{x\to 0}\frac{\sin{3x}}{3x}=3\lim_{z\to 0}\frac{\sin{z}}{z}$. So, what could you say about the last limit? – Salech Alhasov May 10 '12 at 13:22
  • I am not sure, it looks undefined to me. –  May 10 '12 at 13:28
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    One moment there! You said that you know, $\lim_{x\to 0}\frac{\sin{x}}{x}=1$. So, what is the problem? – Salech Alhasov May 10 '12 at 13:28
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    Is it just an accepted fact that this identity is well known? I do not know how to prove it. –  May 10 '12 at 13:34
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    Yes, and in the link below you will find a standard proof of that fact. http://www.ies.co.jp/math/java/calc/LimSinX/LimSinX.html – Salech Alhasov May 10 '12 at 13:38
  • l'hopital rule is easier to use fo this but probably the PO doens't know this (yet ?) – Belgi Jun 01 '12 at 17:52
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In general,

$$\lim_{x\to 0}\frac{\sin{Ax}}{x} = A$$

Rewriting $\lim_{x\to 0}\frac{\sin{Ax}}{x}$ as

$$ A\lim_{x\to 0}\frac{\sin{Ax}}{Ax}$$ (which is legal since an $A$ term would cancel out from the denominator leaving us our original.)

Letting a variable, say, $s = Ax$, we have: $$A\lim_{x\to 0}\frac{\sin{s}}{s}$$

From here, note that as $x$ goes to $0$, so does $s$. Using the well-known fact that $$\lim_{x\to 0}\frac{\sin{x}}{x} = 1$$

We have $$A\cdot1$$ which concludes that $$\lim_{x\to 0}\frac{\sin{Ax}}{x} = A$$

So, your limit is $3.$

Joe
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  • @Belgi At first I felt it was a bit trivial given the comments earlier, but you were right to claim my first answer was a bit brief. :) – Joe Jun 01 '12 at 18:41
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    I felt so too...From reading the PO comments the seems less trivial to him and I felt the if the private case $A=3$ is hard for him the answer won't really help him. Good explenasion – Belgi Jun 01 '12 at 18:42
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Here we'are going to appeal to a very well known inequality:

$$ \sin(x) < x < \tan(x),\space 0<x<\frac{\pi}{2}$$

In your case you have that:

$$ \sin(3x) < 3x < \tan(3x),\space 0<x<\frac{\pi}{6}$$

From the above inequality we get that: $$\cos(3x) < \frac{\sin(3x)}{3x}< 1$$ After multiplying the inequality by 3 and taking the limit when x goes to ${0}$ we get that:

$$\lim_{x\rightarrow0}3\cos(3x) \leq \lim_{x\rightarrow0}\frac{\sin(3x)}{x} \leq 3$$

By Squeeze Theorem the limit is $3$.

The proof is complete.

user 1591719
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  • This doesn't seem quite right ... you might want to add some details or something... – Thomas Jun 01 '12 at 17:15
  • @Thomas: OK. I've just added them. Thanks. – user 1591719 Jun 01 '12 at 17:48
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    Writing $3 < \ldots < 3$ is just wrong. – TMM Jun 01 '12 at 18:40
  • @TMM: i just typed things in a better manner. – user 1591719 Jun 01 '12 at 19:54
  • I wanted to post this solution because it's the most elementary way to solve such a limit. – user 1591719 Jun 01 '12 at 19:59
  • It is still wrong, since $\lim_{x\rightarrow0}3\cos(3x) < \lim_{x\rightarrow0}\frac{\sin(3x)}{x}< 3$ is equivalent to $3 < 3 < 3$. This time I'll include a downvote. – TMM Jun 01 '12 at 20:07
  • @TMM: maybe you wanna see this: http://www.khanacademy.org/math/calculus/v/proof--lim--sin-x--x. Moreover, at school i did things as i presented. It's not fair! – user 1591719 Jun 01 '12 at 20:23
  • @Marvis: i've just received a downvote for writing in my proof what TMM has just typed above. I'd like to have one more opinion. Am i wrong if i use only "<" instead of "<="? In my study book and on other links i visited it is correct. Since it's about a limit then it should be correct. That's why i need one more opinion. – user 1591719 Jun 01 '12 at 20:37
  • @TMM Your downvote is not called for. The proof of the simple case uses $$\cos x< \frac{\sin x} x < 1$$ The use of $\leq$ is superfluous. – Pedro Jun 01 '12 at 22:06
  • @Peter Downvoting is my own decision, and I do not recall asking you for advice. What I said is true and what was written in the answer was wrong, even after the first edit. Chris is too careless with inequality-signs leading to false statements, so I downvoted. If you check the edit history, you will see that in both cases the $\leq$ was not superfluous but necessary. – TMM Jun 01 '12 at 23:46
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Here is another way of looking at this.

\begin{align*} \lim_{x \to 0} \frac{\sin{3x}}{x} &= \lim_{x\to 0} \frac{3\sin(x) - 4 \sin^{3}(x)}{x} \\\ &= 3\cdot \lim_{x \to 0} \frac{\sin{x}}{x} - 4 \cdot \lim_{x \to 0} \frac{\sin{x}}{x} \cdot \lim_{x \to 0} \sin^{2}{x} \\\ &= 3. \end{align*}

You can also expand $\sin(x)$ as a taylor series and then try to get an answer. Note that $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots$ therefore $$\sin(3x) = 3x - \frac{(3x)^{3}}{3!} + \cdots$$ Now just divide the above quantity by $x$ and then take the limit as $x \to 0$.