Let $f \colon [0, \infty) \rightarrow \mathbb{R}$ is given as $f(x) = \left(x^2 + \lfloor x^2\rfloor\right) \sin (2 \pi x)$. Then can we comment on the continuity of $f$?
Here $\lfloor x\rfloor$ is the floor function, or Greatest Integer function.
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Yes, I don't see where $\sin(2\pi x)$ can be non-continuous in $\mathbb{R}$, nor do I find any of that in $x^2$. – JoeyAndres Jul 20 '14 at 14:35
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The second factor is continuous everywhere. So the question is how the function behaves in the discontinuities of the first factor. Where is the first factor discontinuous? – Daniel Fischer Jul 20 '14 at 14:39
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$[x^2]$ is discontinuous for every integer $x$. Am I right? – hola Jul 20 '14 at 14:42
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You are right! This continuous for all $ x\in \mathbb {R}-\mathbb {Z} $ – Mohammad W. Alomari Jul 20 '14 at 14:48
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Did you mean to have $x^2$ in the first terms and $x$ in the $\sin$? That is, did you intend for the last term to be $\sin ( 2 \pi x^2)$? – copper.hat Jul 20 '14 at 14:54
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This function is continuous at each integer because the second term goes to 0 and first term is locally bounded. – Shine Jul 20 '14 at 15:13
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I think the discontinuities=${x\in \mathbb{R}: x^2\in \mathbb{Z}, x\neq \frac{k}{2}, k\in \mathbb{Z}}$. – Shine Jul 20 '14 at 15:16
