<s>[INST] Lim $x\to  0$ of $\frac{\sin(\pi x)}{\tan(\sqrt{3}x)}$ $$
   \lim_{x \to 0} \frac{\sin(\pi x)}{\tan(\sqrt{3} x)}
$$
I need a step by step explanation. Thank you.
 [/INST] Hint:
First, let's expand what we are looking at:
$$\begin{align}
\frac{\sin(\pi x)}{\tan(\sqrt{3}x)} = \frac{\sin(\pi x)}{\frac{\sin(\sqrt{3} x)}{\cos(\sqrt{3} x)}} = \sin(\pi x) \cdot \frac{\cos(\sqrt{3}x)}{\sin(\sqrt{3}x)} = \sin(\pi x)\cdot\frac{1}{\sin(\sqrt{3}x)}\cdot \cos(\sqrt{3}x).
\end{align}$$
Next, we want to get each $\sin$ term in the form $\frac{\sin(t)}{t}$.  One way to do that is to multiply by "$1$" in a helpful way:
$$\begin{align}
\sin(\pi x)\cdot\frac{1}{\sin(\sqrt{3}x)}\cdot \cos(\sqrt{3}x) &= \sin(\pi x)\frac{\pi x}{\pi x} \cdot \frac{1}{\sin(\sqrt{3}x)} \cdot \frac{\sqrt{3}x}{\sqrt{3} x} \cos(\sqrt{3}x)
\\
&= \frac{\sin(\pi x)}{\pi x} \cdot \frac{\sqrt{3} x}{\sin(\sqrt{3} x)} \cdot \frac{\pi x}{\sqrt{3} x} \cos(\sqrt{3} x).
\end{align}$$
I'll let you take a limit and see what happens.
</s>