Evaluating limn→∞1n∫0π2sin2nxsin2xf(x)dx for continuous f
Evaluating for continuous f
Answer & Explanation
Beginner2022-01-30Added 14 answers
The limit is
Since f is continuous on and, hence, bounded we have and and an estimate for the second integral on the RHS of (*) is
Since the second integral converges to 0 for any , it remains to prove that the first integral on the RHS of (*) can be made arbitrarily close to 0 with a suitable choice for .
For any we can chose such that for
Using we get
Since, can be arbitrarily close to 0 we are done.
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