Answer
$$\lim_{x\to0}\frac{\tan(2x)}{\tan(\pi x)}=\frac{2}{\pi}$$
Work Step by Step
$$A=\lim_{x\to0}\frac{\tan(2x)}{\tan(\pi x)}=\lim_{x\to0}\frac{\frac{\sin2x}{\cos2x}}{\frac{\sin\pi x}{\cos \pi x}}=\lim_{x\to0}\frac{\sin2x}{\sin\pi x}\frac{\cos\pi x}{\cos2x}$$
Multiply the nominator with $2x/2x$ and the denominator with $\pi x/\pi x$ and we have
$$A=\lim_{x\to0}\frac{2x\frac{\sin2x}{2x}}{\pi x\frac{\sin\pi x}{\pi x}}\frac{\cos\pi x}{\cos2x}=\lim_{x\to0}\frac{2\cos\pi x\frac{\sin 2x}{2x}}{\pi\cos2x\frac{\sin\pi x}{\pi x}}$$
$$A=\frac{2\cos0\times\lim_{x\to0}\frac{\sin2x}{2x}}{\pi\cos0\times\lim_{x\to0}\frac{\sin\pi x}{\pi x}}=\frac{2\lim_{x\to0}\frac{\sin2x}{2x}}{\pi\lim_{x\to0}\frac{\sin\pi x}{\pi x}}$$
Recall that $\lim_{x\to0}\frac{\sin x}{x}=1$. Here we apply the identity for $2x$ and $\pi x$ instead, with still the same results:
$$A=\frac{2\times1}{\pi\times1}=\frac{2}{\pi}$$
