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I have to evaluate $$\lim_{n\rightarrow\infty}\int^\pi_{\frac{\pi}{2}}\frac{n\mathrm{sin}(\frac{x}{n})}{x}dx.$$ So I think I have to prove that $$\frac{n\mathrm{sin}(\frac{x}{n})}{x}\rightarrow1$$ uniformly. But I can't understand how to estimate $$\sup_{x\in(\frac{\pi}{2},\pi)}\left|\frac{n\mathrm{sin}(\frac{x}{n})}{x}-1\right|.$$ Could you help me, please?

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up vote 1 down vote accepted

More explicitly, since the series for sin is enveloping, $x > \sin x > x-x^3/6$ (or $1 > \sin x/x > 1-x^2/6$) for $0 < x < \pi/2$.

Use this to bound the difference you want.

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how do you prove that $\mathrm{sin}\;x>x-x^3/6$? – Alex M Oct 3 '11 at 4:59
Integrating twice the inequality $\sin x\leqslant x$. – Did Oct 3 '11 at 5:27

Hint: $\lim_{t \to 0} \sin(t)/t = 1$.

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