calculus - Finding $limlimits_{ntoinfty}frac1{n^2}int_0^{frac{pi}2}xleft(frac{sin nx}{sin x}right)^4,mathrm dx$

$$\lim_{n\to\infty}\frac1{n^2}\int_0^{\frac{\pi}2}x\left(\frac{\sin nx}{\sin x}\right)^4\,\mathrm dx$$ I don't know how to solve this limit problem. I'm roughly clear about using the pinch theorem, but I haven't done it yet. Could somebody please tell me? Thanks!

Answer

From

$$\lim_{x\to 0}x^2\left(\frac1{\sin^4 x}-\frac1{x^4}\right)=\frac23,$$
for $x\in(0,\frac\pi 2)$, there is $C$ such that
$$\left|\frac{x\sin^4 nx}{\sin^4 x}-\frac{\sin^4 nx}{x^3}\right|\le \frac{C\sin^4 nx}{x}\le Cn.$$

Thus,

$$\begin{align*} \lim_{n\to\infty}\frac1{n^2}\int_0^{\frac\pi2}x\left(\frac{\sin nx}{\sin x}\right)^4\, \mathrm dx&=\lim_{n\to\infty}\frac1{n^2}\int_0^{\frac\pi2}\frac{\sin^4 nx}{x^3}\, \mathrm dx\\ &= \lim_{n\to\infty}\int_0^{\frac{n\pi}2}\frac{\sin^4 x}{x^3}\, \mathrm dx\\ &= \int_0^{\infty}\frac{\sin^4 x}{x^3}\, \mathrm dx\\ \text{(IBP)}\quad &= \int_0^{\infty}\frac{2\sin^3 x\cos x}{x^2}\, \mathrm dx\\ \text{(IBP)}\quad &= \int_0^{\infty}\frac{6\sin^2 x\cos^2 x-2\sin^4 x}{x}\, \mathrm dx\\ &= \int_0^{\infty}\frac{\cos 2x-\cos 4x}{x}\, \mathrm dx\\ \text{(Frullani)}\quad &=\ln 2. \end{align*}$$