3
$\begingroup$

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.

  • 0
    @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

11

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

  • 0
    l'hopital rule is easier to use fo this but probably the PO doens't know this (yet ?)2012-06-01
3

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
    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
1

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.

  • 0
    @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
1

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 $x \to 0$.