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.

\begin{align}
\int_0^\pi \left|\sin x - \sin(2nx)\right|\mathrm d x &= \frac1{2n}\int_{0}^{2n\pi} \left|\sin\left(\frac{x}{2n}\right) - \sin(x)\right|\mathrm dx\\
&= \frac{1}{2n} \sum_{k=0}^{n-1}\int_{2k\pi}^{2(k+1)\pi}\left|\sin\left(\frac{x}{2n}\right) - \sin(x)\right|\mathrm dx\\
&= \frac{1}{2n}\sum_{k=0}^{n-1}\int_0^{2\pi}\left|\sin(x) - \sin\left(\frac{x+2k\pi}{n}\right)\right|\mathrm d x\\
&\underset {n\to \infty} \to \frac1{4\pi} \int_0^{2\pi}\int_{0}^{2\pi}\left|\sin x - \sin y\right|\mathrm dx\,\mathrm dy\\
&= \frac1{2\pi} \int_0^{\pi}\int_{0}^{\pi}\left(\left|\sin x - \sin y\right| + \left|\sin x + \sin y\right|\right)\mathrm dx\,\mathrm dy\\
&= \frac2{\pi} \int_0^{\frac\pi2}\int_{0}^{\frac\pi2}\left(\left|\sin x - \sin y\right| + \left|\sin x + \sin y\right|\right)\mathrm dx\,\mathrm dy\\
&= \frac2\pi  \int_0^{\frac\pi2}\int_{0}^{\frac\pi2}\left(\left|\sin x - \sin y\right| + \left|\sin x + \sin y\right|\right)\mathrm dx\,\mathrm dy\\
&= \frac4{\pi} \int_0^{\frac\pi2}\left(\int_{0}^{y}\sin y\,\mathrm dx + \int_{y}^{\frac\pi 2}\sin x\,\mathrm d x\right)\mathrm dy\\
&= \frac4\pi\int_0^{\frac\pi2} \left(y\sin y + \cos y\right)\mathrm d y\\
&= \frac4\pi \left[-y\cos y + 2\sin y\right]_{0}^{\frac\pi2}\\
&= \frac8\pi
\end{align}
