Evaluating limn→∞1n∫0π2sin2nxsin2xf(x)dx for continuous f

The limit is $\frac{\pi }{2}f(0)$ Note that $|{\int }_0^{\pi /2}\frac{{\sin }^2(nx)}{n{\sin }^{2}x}f(x)dx-\frac{\pi }{2}f(0)|$ (*) $⩽|{\int }_0^{\delta }\frac{{\sin }^2(nx)}{n{\sin }^{2}x}[f(x)-f(0)]dx|+|{\int }_{\delta }^{\pi /2}\frac{{\sin }^2(nx)}{n{\sin }^{2}x}[f(x)-f(0)],dx|$ Since f is continuous on $[0,\pi /2]$ and, hence, bounded we have $|f(x)|⩽M$ and $|f(x)-f(0)|⩽2M$ and an estimate for the second integral on the RHS of (*) is $|{\int }_{\delta }^{\pi /2}\frac{{\sin }^2(nx)}{n{\sin }^{2}x}[f(x)-f(0)]dx|⩽\frac{2M}{n}{\int }_{\delta }^{\pi /2}\frac{1}{{\sin }^{2}x}dx{\to }_{n\to \mathrm{\infty }}0$ Since the second integral converges to 0 for any $\delta$, it remains to prove that the first integral on the RHS of (*) can be made arbitrarily close to 0 with a suitable choice for $\delta$. For any $ϵ>0$ we can chose $\delta$ such that $|f(x)-f(0)|<ϵ$ for $0⩽x⩽\delta$ Using $\sin x>2x/\pi$ we get $|{\int }_0^{\delta }\frac{{\sin }^2(nx)}{n{\sin }^{2}x}[f(x)-f(0)]dx|⩽\frac{{\pi }^2ϵ}{4n}{\int }_0^{\delta }\frac{{\sin }^2(nx)}{x^2},dx$ $=\frac{{\pi }^2ϵ}{4}{\int }_0^{n\delta }\frac{{\sin }^{2}u}{u^2}du$ $⩽\frac{{\pi }^2ϵ}{4}{\int }_0^{\mathrm{\infty }}\frac{{\sin }^{2}u}{u^2}du=\frac{{\pi }^3ϵ}{8}$ Since, $ϵ$ can be arbitrarily close to 0 we are done.