# Evaluating Limit of an Integral (Real-Analysis)

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!

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Try a trivial bound for the numerator and bound the integral as a function of $n$. –  user17762 Apr 30 '12 at 1:41

Hint: $$\biggl|\,\int_0^1{\sin(n/x)\over n\sqrt x}\,dx\,\biggr|\le {1\over n}\int_0^1{\bigl|\sin(n/x)\bigr|\over \sqrt x}\,dx\le {1\over n}\int_0^1{1\over \sqrt x}\,dx={2\over n}.$$