<s>[INST] 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?
 [/INST] The key is to use Hopital rule:
$$\begin{align}
\lim_\limits{x\to 0}\frac{\sin(x)}{x} &= \lim_\limits{x\to 0}\frac{\sin'(x)}{1}\\
&= \lim_\limits{x\to 0}\frac{\cos(x)}{1}\\
&= \lim_\limits{x\to 0}\frac{1}{1}\\
&=1
\end{align}
$$
hence
$$
\begin{align}
\lim_{x\to 0} \frac{\sin(6x)}{\sin(2x)} &= \lim_{x\to 0} \frac{\sin(6x)}{\sin(2x)} \cdot \frac{2x}{6x} \cdot \frac{6}{2}\\
&= \lim_{x\to 0} \frac{\sin(6x)}{6x} \cdot \frac{\sin(2x)}{6x} \cdot 3\\
&= 1 \cdot 1 \cdot 3\\
&=3
\end{align}$$
</s>