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I have this function $$ f(x)=\sqrt{2-\sqrt{x}} $$

and know that this function is continuous on interval $[0,4]$.

So it means that this function is discontinuous on intervals $(-\infty,0) \cup (4,+\infty)$.

Question: do those intervals represent points of discontinuity of second kind because function is not defined in those points and also limits in those points do not exist?

Am I right? Apologies if this question sounds odd, first time learning continuity of function properly. :)

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    $\begingroup$ We don't consider classification of points where the function has no definition. $\endgroup$ – user9464 Nov 30 '19 at 23:53
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    $\begingroup$ See this: en.wikipedia.org/wiki/Classification_of_discontinuities $\endgroup$ – user9464 Dec 1 '19 at 0:13
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    $\begingroup$ If a function is not continuous at a point in its domain, one says that it has a discontinuity there. $\endgroup$ – user9464 Dec 1 '19 at 0:14
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    $\begingroup$ The function is not $g(x)=\frac{1}{x}$. The function is $f(x)=\begin{cases}\frac1x&x\ne0\\1&x=0\end{cases}$. They are different: the former one has no definition at $x=0$ but the later one does. $\endgroup$ – user9464 Dec 1 '19 at 0:24
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    $\begingroup$ Read in your link this: "... but it is not of the domain of the function, so it is not possible to define the discontinuity." $\endgroup$ – user9464 Dec 1 '19 at 0:25
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We don't consider classification of points where the function has no definition.

See this article : https://en.wikipedia.org/wiki/Classification_of_discontinuities

In particular:

If a function is not continuous at a point in its domain, one says that it has a discontinuity there.

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  • $\begingroup$ OK, now I see that we do not consider classification of points where the function has no definition. Couple minutes ago I read some assignments with solution(school web) and it says something like that: function is continuous on its domain, but we will analyze bulk points of function and then we state that this point is for example of essential discontinuity. I am kinda confused, must admit. :( $\endgroup$ – naruto25 Dec 1 '19 at 0:39
  • $\begingroup$ @naruto25: That would be another question. It may be good if you could post a new one and cite your original assignments. "function is continuous on its domain" is not true. $\endgroup$ – user9464 Dec 1 '19 at 2:21
  • $\begingroup$ Large parts of many areas of analysis are concerned with the boundary behavior of functions and various types of cluster sets of functions (e.g. harmonic analysis, Sobolev classes, complex analysis, partial differential equations, etc.), but of course this is a bit beyond what the OP is asking about, and also the term "discontinuity" is not applied in these situations unless the function is defined at such points on the boundary. $\endgroup$ – Dave L. Renfro Dec 1 '19 at 9:42
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A function $f$ is continuous if is continuous on every point of its domain.

The domain of $\sqrt{2-\sqrt{x}}$ is $[0, 4]$ and the function is continuous in every point of $[0, 4]$ so is a continuous function.

A discontinuity of second kind is a type of irremovable discontinuity such that:

1.The function is not defined only in one side of the point

or

  1. The lateral limits does not exist(one or both sides)

As note: If the function is undefined in both sides of a point $x_0$ , there is not a discontinuity of second type. A function is always continuos in isolated points.

You can read more about this in wikipedia.

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