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?
The limit of an integral
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real-analysis
analysis
2 Answers
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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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2Integrating twice the inequality $\sin x\leqslant x$. – 2011-10-03
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Hint: $\lim_{t \to 0} \sin(t)/t = 1$.