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In my book, there is a sentence that says exactly this:

"The function $\mathrm{sgn}(x)= \dfrac{x}{|x|}$ is neither continuous nor discontinuous at $x=0$. How is this possible?"

It was easy for me to tell it is not continuous at $x=0$ as there is no limit existence due to left-right limit inequality, or simply because the graph is broken at $x=0$. But I can't understand the second part, which claims it is also not discontinuous.

I've always thought that "if it is not continuous, then discontinuous", but apparently it seems to be wrong. How is this function not discontinuous?

source: A complete course: calculus(8th edition)

user2694307
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  • The quote I wrote means it is not discontinous, doesn't it? I am asking this to be sure if I got the meaning right. – user2694307 Oct 04 '14 at 12:44
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    The function is undefined at $x=0$, maybe your book requires the function to at least be defined at the point to be able to even speak about continuity/discontinuity? – Raskolnikov Oct 04 '14 at 12:44
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    @Raskolnikov nailed it. That is indeed standard usage. (But normally, one defines $\operatorname{sgn}(0)=0$.) – Harald Hanche-Olsen Oct 04 '14 at 12:48
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    @Raskolnikov, I've come across an example of y=(1/x) in the book right now. Book says it is undefined at x= 0, just like you said, giving an extra information it would be discontinous to some authors. – user2694307 Oct 04 '14 at 12:57
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    Related discussions: http://math.stackexchange.com/q/1087623/1242, http://math.stackexchange.com/q/1449507/1242. – Hans Lundmark Nov 26 '15 at 10:56
  • Please can you send me the picture where it is written that the function is not continuous and discontinuous at the same time ! ? – Heena Aug 23 '17 at 09:27
  • From my understanding which is the same as Wikipedia signum funciton, the signum function is defined at zero which is unlike the definition you quoted. The quoted function is almost the same as signum. In any case, a function can be continuous or discontinuous only where it is defined. – Somos Feb 13 '21 at 22:32

4 Answers4

1

A function cannot be continuous and discontinuous at the same point.

Yes the function is discontinuous which is right as per your argument.

I think the question wanted to convey this..

It has a jumped discontinuity which means if the function is assigned some value at the point of discontinuity it cannot be made continuous.

But the function is definitely discontinuous at x=0.

Jasser
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  • But it is continuous on $\Bbb{R}-{0}$. – Marc Bogaerts Oct 04 '14 at 15:41
  • I mean to say the function cannot be continuous and discontinuous at the same point. @Nimda – Jasser Oct 04 '14 at 16:32
  • So is the function $f(x)=x^3$. It is continuous on $\Bbb{R}$ but not on the one point compactification of $\Bbb{R}$. – Marc Bogaerts Oct 04 '14 at 16:34
  • Sorry I am not very good in understanding big words so please describe one point compactification and what I know is that polynomials are continuous on the whole real number set. – Jasser Oct 04 '14 at 16:38
  • See this link http://en.wikipedia.org/wiki/Compactification_(mathematics)#Alexandroff_one-point_compactification – Marc Bogaerts Oct 04 '14 at 16:40
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    The function $f(x)=x/|x|$ is continuous according to the standard definition, meaning "continuous at every point of its domain of definition". In order to avoid the somewhat strange situation of having to say that a continuous function has a discontinuity, many books (including, apparently, the one referred to in the question) reserve the word "discontinuous" for points where $f$ is defined. With this definition, a function can neither be continuous nor discontinuous at a point where it's undefined. – Hans Lundmark Nov 26 '15 at 11:01
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the sgn function is a discontinuous function (isolated/jump discontinuity). it will be continuous if domain of the function is changed to R- or R+.

You can get some info about the types of discontinuity from the following sources: https://en.wikipedia.org/wiki/Continuous_functio https://en.wikipedia.org/wiki/Classification_of_discontinuities#Removable_discontinuity

Jesdin Raphael
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The function is neither continuous nor discontinuous at x=0 as it is not defined at x=0 . It simply does not have a value at x=0 .

Singhmeheh
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-2

The function sgn$(x)$ is undefined at $x=0$. A function can be continuous or discontinuous at the points of domain not at points outside of domain. Since $\{0\}$ does not belong to the domain of sgn$(x)$, the question of continuity or discontinuity is irrelevant.

However limit does not at the $N(0)-\{0\}$ so it is neither continuous nor discontinuous.

dantopa
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Shourja Chowdhury
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