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Say I have a function of the form $$\frac{x+3}{x^2 - 9 }$$ this can be factorized, and reduced to: $$\frac{1}{x-3}$$ In the first form, the function is discontinuous at $x = \pm 3 $. However the second form, is only discontinuous at $x = 3$. Does that mean that, the function is discontinuous only at x = 3 ? I have read other posts here that said evaluating functions at undefined points does not make sense, but then how does one test for continuity?

RMS
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2 Answers2

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In the first form the function is not defined at $x = \pm 3$. Those points are not in the domain. The function is continuous on its domain. The second form is the same function as the first where both are defined, and adds the point $x=-3$ to the domain in such a way as to make the new function continuous on its (larger) domain. That's what it means to "remove the singularity" at $x=-3$. There is no way to remove the singularity at $x=3$ that will create a function continuous there.

Ethan Bolker
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  • In many math books and online examples, whenever I am asked to find the "discontinuities" of a function, they always end up being the undefined points? Why is that? Is it an oversimplification? I am getting really confused. – RMS Feb 20 '23 at 17:49
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    The discussions in those "many math books" are a little sloppy/oversimplified to make the authors' points in shorter sentences than would be required to get the words exactly right. I think that's usually OK. Now that you know the full story you can fill in the right words when you see these discussions. – Ethan Bolker Feb 20 '23 at 18:08
  • But it is a fair statement to say the function is not continuous at a point P, that lies out of its domain, right? – RMS Feb 22 '23 at 20:15
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    @RSM It's literally not correct. To state the definition of continuity you need the value of the function at the point. Whether or not it's "fair" to elide that is a matter of opinion. The subtlety is what led to your good question in the first place. – Ethan Bolker Feb 22 '23 at 21:20
  • at that point the condition for continuity is not satisfied since the function is undefined, so the function is discontinuous at that point. By fair, I meat “valid”, my bad ! – RMS Feb 23 '23 at 06:23
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I hope this answers your confusion.

Discontinuities are points where the function is not continuous. Given a function $f(x)$, it is continuous at point $a$ if $$\lim_{x\to a} f(x)=f(a)$$

So a discontinuity is a point simply not satisfying this. If a function is undefined, it in particular does not satisfiy the condition, since the right hand side is undefined and therefore can't be equal to anything. The function in question is undefined for $x=\pm 3$, causing $x=\pm 3$ to be discontinuities.

HappyDay
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