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Wolfram tells me that for any $\alpha \ge 1$ (I tried up to $\alpha = 10^8$). $$\lim_{x \rightarrow 0} \frac {\sin \left(\frac 1x\right)}{x^\alpha} < \infty$$ How can I show that this is true?

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  • $\begingroup$ It should be $\alpha<0$! $\endgroup$ – Mhenni Benghorbal Feb 15 '16 at 4:59
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    $\begingroup$ There's no way this converges for $\alpha\geq 0$... $\endgroup$ – bartgol Feb 15 '16 at 5:00
  • $\begingroup$ Look at the plot that wolframalpha makes. It clearly shows something fishy is going on around 0. If you try a small $\alpha$ (1 or 2) you clearly see the mess that is going on around 0... $\endgroup$ – bartgol Feb 15 '16 at 5:04
  • $\begingroup$ I include the plot here too. It only shows the imaginary part. $\endgroup$ – Paichu Feb 15 '16 at 5:06
  • $\begingroup$ Have you seen my Comment? $\endgroup$ – Mhenni Benghorbal Feb 15 '16 at 5:35
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Function $\sin(1/x)$ is bounded as $x\rightarrow0$, while $x^{-\alpha} \rightarrow 0$ for $\alpha<0$. As a result, the limit of their product is $0$ as well.

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  • $\begingroup$ The limit in the question is for $x\to 0$... $\endgroup$ – bartgol Feb 15 '16 at 5:01
  • $\begingroup$ @bartgol: my bad. $\endgroup$ – Pavel Feb 15 '16 at 5:02

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