Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$? Why is $\lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = \frac{6}{2}=3$?
The justification is that $\lim_\limits{x\to 0}\frac{\sin(x)}{x} = 1$
But, I am not seeing the connection.
L'Hospital's rule? Is there a double angle substitution happening?

In asymptotic terms, $\displaystyle \lim_\limits{x\to 0}\frac{\sin(x)}{x} = 1$ means that "$\sin(x)$ behaves like $x$ around $0$". 
To this understanding, $\sin(6x)$ behaves like $6x$ and $\sin(2x)$ behaves like $2x$ around $0$. 
Therefore, $\displaystyle \frac{\sin(6x)}{\sin(2x)}$ behaves like $\displaystyle \frac{6x}{2x}=3$ around $0$. 
In other words,  $\displaystyle \lim_\limits{x\to 0}\frac{\sin(6x)}{\sin(2x)} = 3.$
