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
Question
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
Solution
To prove that fn ∈ R[0, 1] for all n ∈ N, we can use the fact that fn is continuous on [0, 1]. This can be shown by using the inequality |sin(x^2n)| ≤ x^2n for x ∈ [0, 1]. Therefore, fn is continuous on [0, 1] and belongs to R[0, 1] for all n ∈ N.
To prove that limn→∞Z 10fn = 0, we can use the comparison principle. For all n ≥ N, we have Z 10fn ≤ Z 10|fn| ≤ Z 10xn = 1/n + 1.
By applying the squeeze theorem, we can conclude that limn→∞Z 10fn = 0.
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