Evaluate the limit $\lim\limits_{n\rightarrow \infty}\int_{0}^{\pi}\left|\sin(x)-\sin(2nx)\right|\text{ d}x$ 
$$\lim_{n\rightarrow \infty}\int_{0}^{\pi}\left|\sin(x)-\sin(2nx)\right|\mathrm d x$$

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I put proper big numbers to $n$ using calculator to estimate the answer.
And then I figured out that the answer seems like $\frac{8}{\pi}$, but
unfortunately I dont have any clues.
Please help.

Hint:
Split the interval $I = [0,\pi]$ to:
$$I = \bigcup\limits_{k=0}^{2n-1} I_k$$
where $I_k = \left[\dfrac{\pi k}{2n}, \dfrac{\pi (k+1)}{2n}\right).$ You should be easily able to calculate a closed formula for:
$$J_k = \int_{I_k}|\sin x - \sin 2nx|dx$$
because the choice of $I_k$ restricts $2nx\in [\pi k, \pi (k+1)).$ For instance, if you have an odd $k$, the absolute value can be forgotten because $\sin x\geq 0$ while $\sin 2nx\leq 0.$ For even $k$, you just need to do a little more work.
