<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] As user Daniel has written you can express the ratio $\frac{\sin 6x}{\sin 2x}$ in a way which makes it amenable to the use of standard limit $$\lim_{x\to 0} \frac{\sin x}{x} = 1\ .$$ Notice that $$\lim_{x\to 0} \frac{\sin x}{x} = \lim_{x\to 0} \frac{\sin 2x}{2x} = \lim_{x\to 0} \frac{\sin 6x}{6x}\ .$$ As long as the argument $x$ is not equal to zero (and $\sin 2x \neq 0$) you can prolong the ratio $\frac{\sin 6x}{\sin 2x}$ as follows.
$$\frac{\sin 6x}{\sin 2x} = \frac{\sin 6x}{6x} \cdot \frac{6x}{2x} \cdot \frac{2x}{\sin 2x} = \underbrace{\frac{\sin 6x}{6x}}_{\to 1} \cdot \frac{6}{2} \cdot \underbrace{\frac{1}{\frac{\sin 2x}{2x}}}_{\to\frac{1}{1}} \to 1 \cdot 3 \cdot 1 = 3 \ .$$ Hence you obtain the result $$\lim_{x\to 0}\frac{\sin 6x}{\sin 2x} = 3 \ .$$
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