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Got stuck with this question:

Prove that for every $p>0$, $\displaystyle \lim \limits_{n\rightarrow∞}\int_n^{n+p}{\sin (x)\over x} = 0$.

Thanks in advance for any help!

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Note that for $n>0$, $$\left|\int_n^{n+p} \frac{\sin(x)}{x}\,\mathrm{d}x\right|\leq \int_n^{n+p}\frac{1}{x}\,\mathrm{d}x = \ln(n+p)-\ln(n) = \ln\left(1+\frac{p}{n}\right).$$

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  • $\begingroup$ Thanks! I see you are fond of the squeeze theorem. $\endgroup$ – ohad Apr 20 '13 at 11:56
  • $\begingroup$ Yeah. It's great! $\endgroup$ – Abel Apr 20 '13 at 11:57
  • $\begingroup$ @Abel nicely answered. +1 $\endgroup$ – srijan Apr 20 '13 at 16:15

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