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Prove thatlimn→∞Z 10fn = 0.Hint) Use the comparison principle

Question

Prove thatlimn→∞Z 10fn = 0.Hint) Use the comparison principle

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Solution

I'm sorry, but your question seems to be incomplete. Could you please provide the full details or the function of 'fn'? The comparison principle in limits states that if 0 ≤ fn ≤ gn for all n and the limit of gn as n approaches infinity is 0, then the limit of fn as n approaches infinity is also 0. However, without the full details of 'fn', I can't provide a step-by-step proof.

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(11-12) For each n ∈ N, define fn : [0, 1] → R byfn(x) =(xn cos(1/x) if x > 0,0 if x = 0.11. Prove that fn ∈ R[0, 1] for all n ∈ N.12. Prove thatlimn→∞Z 10fn = 0.Hint) Use the comparison principle

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For each n ∈ N, define fn : [0, 1] → R byfn(x) =sin(x2n) cos(x2n)xn if x > 0,0 if x = 0.Prove that fn ∈ R[0, 1] for all n ∈ N and thatlimn→∞Z 10fn = 0.Proof. For each n ∈ N, using the inequality | sin(x2n)| ≤ x2n for x ∈ [0, 1], one sees that fn iscontinuous on [0, 1]. Hence, fn ∈ R[0, 1] for all n ∈ N. Further, by the comparison principle, for alln ≥ N,Z 10fn ≤Z 10|fn| ≤Z 10xn = 1n + 1 .Thus, by the squeeze theorem, we obtainlimn→∞Z 10fn = 0

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