calculus - Not sure how to evaluate $lim_{xto 0}frac{sin6x}{sin2x}$ (without l'Hospital)

$$\lim_{x\to 0}\frac{\sin(6x)}{\sin(2x)}$$

I know I can use L'Hospital's but I want to understand this particular explanation. They seem to skip something, and I'm not seeing the connection:

The limit is $\frac{6}{2}=3$ since $\lim_{x\to 0}\frac{\sin(x)}{x}=1$

Answer

For $x \neq 0$ and $x$ close to zero, we have

$$\frac{\sin 6x}{\sin 2x} = \frac{\sin 6x}{6x} \cdot \frac{6x}{\sin 2x} = 3 \cdot \frac{\sin 6x}{6x} \cdot \frac{2x}{\sin 2x}$$

See what to do now?