# Finding $\lim\limits_{x\to 0}\frac{\sin{3x}}{x}$

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.

-
Can you use what you know to figure out $\lim_{n\to\infty}(\sin3x)/(3x)$? – Gerry Myerson May 10 '12 at 13:12
@Salech, just so you know: it's a bad idea to have titles that are entirely in $\LaTeX$. Please avoid them. – J. M. May 10 '12 at 13:18
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
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
@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

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

-
I do not see what that does for the problem. – user138246 May 10 '12 at 13:15
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. – user138246 May 10 '12 at 13:28
Is it just an accepted fact that this identity is well known? I do not know how to prove it. – user138246 May 10 '12 at 13:34
Yes, and in the link below you will find a standard proof of that fact. ies.co.jp/math/java/calc/LimSinX/LimSinX.html – Salech Alhasov May 10 '12 at 13:38

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 $3.$

-
+1 for the edited version! – Belgi Jun 1 '12 at 18:39
@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 1 '12 at 18:41
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 1 '12 at 18:42

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 $x \to 0$.

-

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.

-
This doesn't seem quite right ... you might want to add some details or something... – Thomas Jun 1 '12 at 17:15
@Thomas: OK. I've just added them. Thanks. – user 1618033 Jun 1 '12 at 17:48
Writing $3 < \ldots < 3$ is just wrong. – TMM Jun 1 '12 at 18:40
@TMM: i just typed things in a better manner. – user 1618033 Jun 1 '12 at 19:54
I wanted to post this solution because it's the most elementary way to solve such a limit. – user 1618033 Jun 1 '12 at 19:59