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I was watching a video on the main continuity theorem, and the following slide came up:

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Even though $f(x) = \sin \frac{1}{x}$ is not continuous at $x = 0$, we could still say it is continuous, or continuous on its domain, correct? Since its domain doesn't include $x = 0$.

Chrisuu
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    Yes, we can say that. – Botond May 23 '19 at 17:58
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    It's also not discontinuous at $x=0$, for the same reason. :-) Cf. this question: https://math.stackexchange.com/questions/1087623/is-function-f-mathbb-c-0-rightarrow-mathbb-c-prescribed-by-z-rightarrow – Hans Lundmark May 23 '19 at 20:31

4 Answers4

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Yes

An example illustrating this for an isolated point is $$f(x)=\sqrt{x^4-x^2}$$ which has domain $(-\infty,-1] \cup \{0\} \cup [1, \infty)$.

This $f(x)$ is continuous on its domain according to the definition of continuity, even at the isolated point $x=0$, as there are no other points near enough to $0$ to demonstrate discontinuity

Henry
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Yes. Since the assertion “$f$ is continuous” means “for each $a$ in the domain of $f$, $f$ is continuous at $a$”.

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Yes, that's right. $f(x)$ is continuous on every point in its domain; there are no discontinuities where the function is defined.

auscrypt
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If you define a function to be continuous iff it is continuous at every point in its domain, then yes. However, this can cause confusion. I think it's best to always specify where a function is continuous, whether it be the entire domain or some subset of it.

JDZ
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