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. 
 A: 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 $3.$
A: Hint: $\dfrac{\sin 3x}{x}=3\dfrac{\sin 3x}{3x}$
A: 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.
A: 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 $x \to 0$.
