Answer
$$\frac{2}{\pi}$$
Work Step by Step
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\tan 2 x}{\tan \pi x}&=\lim _{x \rightarrow 0} \frac{\sin 2 x}{\cos 2 x} \cdot \frac{\cos \pi x}{\sin \pi x}\\
&=\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{2 x}\right)\left(\frac{\cos \pi x}{\cos 2 x}\right)\left(\frac{\pi x}{\sin \pi x}\right)\left(\frac{2 x}{\pi x}\right)\\
&=1 \cdot 1 \cdot 1 \cdot \frac{2}{\pi}\\
&=\frac{2}{\pi}
\end{aligned}
