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?

Taking a look at this page we find
$$1+2\cos u + 2 \cos 2 u + 2 \cos 3 u + ... + 2 \cos n u=\frac{\sin \frac{2n+1}{2}u}{\sin\frac u2}$$
In the left hand side set $u=4x$ and $n=1$ so that the left hand side becomes $$\frac{\sin 6x}{\sin 2x},$$ therefore we have $$1+2\cos 4x = \frac{\sin 6x}{\sin 2x}$$
