Let $f :\mathbb{R}\to \mathbb{R}$ be a continuous function with period $1$ and $$\lim_{n\to\infty}\int_0^1\sin^2(\pi x)f(nx)dx= \frac{1}{k}\int_0^1f(x)dx.$$ Find $k$.
My approach till now:
Applying half angle formula $2\sin^2(x) = 1-\cos(2x)$ I got : $$\frac{1}{2}\int_0^1f(nx)dx- \frac{1}{2}\int_0^1\cos(2\pi x)f(nx)dx.$$
I can't think of a way forward from here without applying integration by parts but i don't know if its right to apply it as we don't know about the differentiability of the function $f$.
Please help me with this problem.
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