Evaluating \lim_{n\to\infty}\frac1{n}\int_{0}^{\pi /2}\frac{\sin^2nx}{\sin^2 x}f(x)dx for continuous f



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Evaluating limn1n0π2sin2nxsin2xf(x)dx for continuous f

Answer & Explanation

Emilie Booker

Emilie Booker

Beginner2022-01-30Added 14 answers

The limit is π2f(0) Note that |0π/2sin2(nx)nsin2xf(x)dxπ2f(0)| (*) |0δsin2(nx)nsin2x[f(x)f(0)]dx|+|δπ/2sin2(nx)nsin2x[f(x)f(0)],dx| Since f is continuous on [0,π/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 |δπ/2sin2(nx)nsin2x[f(x)f(0)]dx|2Mnδπ/21sin2xdxn0 Since the second integral converges to 0 for any δ, it remains to prove that the first integral on the RHS of (*) can be made arbitrarily close to 0 with a suitable choice for δ. For any ϵ>0 we can chose δ such that |f(x)f(0)|<ϵ for 0xδ Using sinx>2x/π we get |0δsin2(nx)nsin2x[f(x)f(0)]dx|π2ϵ4n0δsin2(nx)x2,dx =π2ϵ40nδsin2uu2du π2ϵ40sin2uu2du=π3ϵ8 Since, ϵ can be arbitrarily close to 0 we are done.

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