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this may sound like a stupid question, so please excuse me for that. From my understanding after quickly reading up on it (only know the definition of general limits before), $\limsup\limits_{x\rightarrow\infty} sin(x) = 1$. Because $f \in O(g) \leftrightarrow \limsup\limits_{x\rightarrow\infty} |\frac{f(x)}{g(x)}| < \infty$ I can state that $f(x) = 0.5 \in O(sin(x))$.

My problem is when looking at the "regular" Big-O Definition, I am not really sure whether my conclusion $f(x) = 0.5 \in O(sin(x))$ is correct as $sin(x)$ oscillates between 0 and 1, thus there is no $c$ for that $0.5 < c \cdot sin(x) \forall n > n_0$. Those two results contradict and I do not know where my error is...

I would be glad if somebody should tell me where I did something wrong.

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    $\begingroup$ Your feeling about this seems right. Where did you get the equivalent formulation for $O(g)$? That does not look right. $\endgroup$ – 4-ier Aug 26 '18 at 21:57
  • $\begingroup$ @4-ier I got it from en.wikipedia.org/wiki/… (2. row of the table) $\endgroup$ – Maxbit Aug 26 '18 at 22:01
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    $\begingroup$ Are you sure that $\limsup\limits_{x\rightarrow\infty} |\frac{0.5}{\sin(x)}| < \infty$? $\endgroup$ – Mjiig Aug 26 '18 at 22:01
  • $\begingroup$ Yup. Completely agree with @Mjiig $\endgroup$ – 4-ier Aug 26 '18 at 22:03
  • $\begingroup$ @Mijig Ohh thank you. I visualized $0.5 / sin(x)$ wrongly in my head. When looking at a plot, of course it's not $< \infty$. Thank you! $\endgroup$ – Maxbit Aug 26 '18 at 22:06

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