Evaluating $\lim_{x \to 1} \frac{\sin{\pi x}}{\sin{3\pi x}}$ without L'Hopital's rule [closed]

I need help finding this limit:

$$\lim_{x \to 1} \frac{\sin{\pi x}}{\sin{3\pi x}}$$

I've thought of dividing and multiplying the numerator by $\pi x$ and the denominator by $3\pi x$ but it doesn't seem to get me anywhere.

Note that I'm not really familiar with L'Hopital's rule as of now, so any answers which do not include it would be much appreciated.

Recall that

$$\sin(A+B)=\sin A\cos B+\cos A\sin B$$

We have

\begin{align*} \lim_{x \to 1} \frac{\sin{\pi x}}{\sin{3\pi x}}&=\lim_{x \to 1} \frac{\sin{\pi x}}{\sin(2\pi x+\pi x)}\\ &=\lim_{x \to 1} \frac{\sin{\pi x}}{\color{red}{\sin(2\pi x)}\cos(\pi x)+\cos(2\pi x)\sin(\pi x)}\\ &=\lim_{x \to 1} \frac{\sin{\pi x}}{\color{red}{2\sin(\pi x)\cos(\pi x)}\cos(\pi x)+\cos(2\pi x)\sin(\pi x)}\\ &=\lim_{x \to 1} \frac{1}{2\cos(\pi x)\cos(\pi x)+\cos(2\pi x)}\\ &=\frac{1}{2\cos(\pi)\cos(\pi)+\cos(2\pi)}\\ &=\frac13 \end{align*}