Evaluating $\lim_{n\to\infty}\frac1{n}\int_{0}^{\pi /2}{\frac{\sin^2nx}{\sin^2 x}f(x)\;dx}$, for continuous $f$ Let $f:\left[ 0,{\pi }/{2}\; \right]\to \mathbb{R}$ be a continuous function. What is the value of the following limit?

$$\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\int_{0}^{{\pi }/{2}\;}{\frac{{{\sin }^{2}}\left( nx \right)}{{{\sin }^{2}}\left( x \right)}f\left( x \right)dx}$$

I strongly believe that the solution depends on this integral
$$\int_{0}^{{\pi }/{2}\;}{\frac{{{\sin }^{2}}\left( nx \right)}{{{\sin }^{2}}\left( x \right)}dx}=\frac{n\pi }{2}$$
and using Weierstrass Approximation Theorem.
 A: The limit is $\displaystyle \frac{\pi}{2}f(0)$.
Note that
$$\tag{*} \left|\int_0^{\pi/2} \frac{\sin^2(nx)}{n\sin^2 x}f(x) \, dx - \frac{\pi}{2}f(0)\right| \\\leqslant  \left|\int_0^{\delta} \frac{\sin^2(nx)}{n\sin^2 x}[f(x)-f(0)] \, dx\right|+ \left|\int_\delta^{\pi/2} \frac{\sin^2(nx)}{n\sin^2 x}[f(x)-f(0)] \, dx\right|$$
Since $f$ is continuous on $[0,\pi/2]$ and, hence, bounded we have $|f(x)| \leqslant M$ and $|f(x) - f(0)| \leqslant 2M$ and an estimate for the second integral on the RHS of (*) is
$$\left|\int_\delta^{\pi/2} \frac{\sin^2(nx)}{n\sin^2 x}[f(x)-f(0)] \, dx\right| \leqslant \frac{2M}{n}\int_\delta^{\pi/2} \frac{1}{\sin^2 x} \, dx \to_{n \to \infty} 0$$
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 $\epsilon >0$ we can chose $\delta$ such that  $|f(x) - f(0)| < \epsilon$ for $0 \leqslant x \leqslant \delta$.  
Using $\sin x > 2x/\pi$ we get 
$$\left|\int_0^{\delta} \frac{\sin^2(nx)}{n\sin^2 x}[f(x)-f(0)] \, dx\right| \leqslant \frac{\pi^2 \epsilon}{4n} \int_0^\delta \frac{\sin^2 (nx)}{x^2} \, dx \\ = \frac{\pi^2 \epsilon}{4} \int_0^{n\delta} \frac{\sin^2 u}{u^2} \, du \\ \leqslant \frac{\pi^2 \epsilon}{4} \int_0^{\infty} \frac{\sin^2 u}{u^2} \, du  = \frac{\pi^3 \epsilon}{8}$$
Since, $\epsilon$ can be arbitrarily close to $0$ we are done.
