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I have been assigned to investigate the continuity of the function $f(x) = \frac{1}{\sqrt{x}}$. I do know that the function $f$ possesses a discontinuity at $x=0$, however I also know that it is undefined $\forall x < 0$. Should I say that it is discontinuous at those points as well (for the function is undefined) or should I not include them into my final answer, i.e. the set of discontinuities and their classification?

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  • $\begingroup$ The question is a bit vague and lacking details. If a function is not defined in some points, i.e. points lay outside the function's domain, it makes no sense to talk about continuity in those points. What you could do is trying to extend the domain in a way that it preserves continuity in those points. $\endgroup$
    – Philipp
    Sep 27, 2021 at 12:03
  • $\begingroup$ I mean, the function I mentioned is not defined, say, at x = -1. Should I also say that it is discontinuous at that point or it makes no sense to even consider because the function is undefined in the neighbourhood of it? $\endgroup$
    – Barbatulka
    Sep 27, 2021 at 12:08
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    $\begingroup$ The assignment should begin like this. The domain of $f$ is the interval $(0,\infty)$. It makes no sense to discuss continuity for points not in the domain, so I will consider only points in $(0,\infty)$. $\endgroup$
    – GEdgar
    Sep 27, 2021 at 12:13
  • $\begingroup$ How do you know the function has a discontinuity at $x=0$? It seems you have already assumed an answer to your own question. Or did you think $x=0$ is somehow in the domain of $1/\sqrt x$? $\endgroup$
    – David K
    Sep 27, 2021 at 12:26
  • $\begingroup$ @DavidK That is exactly what I cannot understand. The textbook says that the function, say, 1/x possesses an infinite discontinuity at x = 0, which is out of the function’s domain. Why can we even talk about the continuity of a function at the points which are outside of its domain? $\endgroup$
    – Barbatulka
    Sep 27, 2021 at 12:56

1 Answer 1

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The given function is continuous on its natural domain that is $x\in(0,\infty)$. Assuming, for example, $f(0)=0$, this new function would be discontinuous at $x=0$ since $\lim_{x\to 0^+} f(x)\neq f(0)$.

Refer also to the related

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