I'm trying to find $\lim_{n\to\infty}\int_0^1 \frac{\sin(\frac{n}{x})}{n\sqrt{x}}dx\;.$ I know it equals $0$, but I'm having trouble proving why. Help would be greatly appreciated!
Evaluating Limit of an Integral (Real-Analysis)
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real-analysis
integration
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0Try a trivial bound for the numerator and bound the integral as a function of $n$. – 2012-04-30
1 Answers
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Hint: $\biggl|\,\int_0^1{\sin(n/x)\over n\sqrt x}\,dx\,\biggr|\le {1\over n}\int_0^1{\bigl|\sin(n/x)\bigr|\over \sqrt x}\,dx\le {1\over n}\int_0^1{1\over \sqrt x}\,dx={2\over n}.$