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. 

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 $3.$
