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.
 A: 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.
A: Results going to be used:


*

*$\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x} =1 = \lim_{x \to 0} \frac{\tan{x}}{x}$


What you have is 
\begin{align*}
\lim_{x \to 0}\: \frac{\sin(\pi{x})}{\tan\sqrt{3}{x}} &= \lim_{x \to 0} \: \frac{\sin\pi{x}}{\pi{x}} \times \frac{\sqrt{3}x}{\tan{\sqrt{3}x}} \times \frac{\pi{x}}{\sqrt{3}x} \\ &= \lim_{x \to 0} \: \frac{\pi{x}}{\sqrt{3}x} = \frac{\pi}{\sqrt{3}}
\end{align*}
A: As you have a $0/0$ form, L'Hôpital's rule works nicely here:
$$\eqalign{
\lim_{x\rightarrow 0}{\sin(\pi x)\over \tan(\sqrt3 x) }
&=\lim_{x\rightarrow 0}{\bigl[ \sin(\pi x)\bigr]'\over \bigl[\tan(\sqrt3 x)\bigr]' }\cr
&=\lim_{x\rightarrow 0}{  \cos(\pi x) \cdot [\pi x]'\over  \sec^2(\sqrt3 x)\cdot\bigl[ \sqrt 3 x\bigr]' }\cr
&=\lim_{x\rightarrow 0}{\pi\cos(\pi x)\over \sqrt 3\,\sec^2(\sqrt3 x) }\cr
&=\lim_{x\rightarrow 0}\Bigl[{\pi \over \sqrt 3}\cos(\pi x)\,\cos^2(\sqrt3 x) \Bigr]\cr 
&= {\pi\over\sqrt3}.}
$$
Of course, the other answers are nicer, as they are more elementary.
