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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.

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    Can you use what you know to figure out $\lim_{n\to\infty}(\sin3x)/(3x)$?2012-05-10
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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.2012-05-10
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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.2012-05-10
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    Though, in Firefox at least, pressing the shift key and right clicking will pop up the usual context window.2012-05-10
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    @David: still an additional inconvenience, so we prefer that titles not be entirely in $\LaTeX$.2012-05-10
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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).2012-05-10
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    Do you know L'Hospital's rule?2012-05-10
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    Yes but I am not sure if it is suppose to be used at this point in the book.2012-05-10
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    You are probably right.2012-05-10
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    @MarkDominus: I wonder, how do you find the derivative of $\sin$ to use L'Hopital's rule.2012-05-10
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    @Ilya [Booyah](http://math.stackexchange.com/questions/117124/is-using-lhospitals-rule-to-prove-lim-limits-x-to-0-sinx-x-circular/117180#117180)2012-05-10
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    @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$2012-05-10
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    @Ilya I'm not saying it is easier, only that it is possible. =) And you're certainly right on that last remark.2012-05-10

4 Answers 4

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

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    I do not see what that does for the problem.2012-05-10
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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?2012-05-10
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    I am not sure, it looks undefined to me.2012-05-10
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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?2012-05-10
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    Is it just an accepted fact that this identity is well known? I do not know how to prove it.2012-05-10
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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.html2012-05-10
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    l'hopital rule is easier to use fo this but probably the PO doens't know this (yet ?)2012-06-01
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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 $3.$

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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. :)2012-06-01
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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 explenasion2012-06-01
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Here we'are going to appeal to a very well known inequality:

$$ \sin(x) < x < \tan(x),\space 0

In your case you have that:

$$ \sin(3x) < 3x < \tan(3x),\space 0

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.

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    This doesn't seem quite right ... you might want to add some details or something...2012-06-01
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    @Thomas: OK. I've just added them. Thanks.2012-06-01
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    Writing $3 < \ldots < 3$ is just wrong.2012-06-01
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    @TMM: i just typed things in a better manner.2012-06-01
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    I wanted to post this solution because it's the most elementary way to solve such a limit.2012-06-01
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    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.2012-06-01
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    @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!2012-06-01
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    @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.2012-06-01
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    @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.2012-06-01
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    @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.2012-06-01
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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 $x \to 0$.