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This is a follow-up to a question asked by a calculus student here: Is the function $\frac{x^2-x-2}{x-2}$ continuous?

It got me thinking about a more interesting related question in classical analysis: When in general is a discontinuity for a real valued map removable? A rational function with a factorable polynomial in the numerator with a factor in the denominator is an almost trivial case - but when is it true in general? In complex analysis, for holomorphic functions, there's a very well-defined theory based on a theorem of Riemann. But what about real valued maps?

While I'm on the subject, I should note that "removable discontinuity" in the sense used in this problem is something of a misnomer - the point really should be called a removable singularity. The technical distinctions are nicely summed up here:

http://en.wikipedia.org/wiki/Classification_of_discontinuities

Addendum: I just found this older,related post at this board: Is there a function with a removable discontinuity at every point? If this is a valid algorithm, it may provide a starting point for a theoretical basis for the solution to my question!

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When the left and right limits of the function at the singularity exist and agree? –  user7530 Nov 26 '11 at 8:33
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@Mathemagician1234: Your comment was off-topic. Please either calmly ask "Why the downvote?", or don't say anything about it. –  Zev Chonoles Nov 26 '11 at 8:57
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I wasn't the first downvote, but I just added a second downvote. Given your self-reported background, this question is really silly and trivial. A version of it appears as an exercise in every textbook I've ever used to teach freshman calculus. –  Adam Smith Nov 26 '11 at 21:21
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@Mathemagician1234: Your comment does not seem to me to make a lot of sense; the two-sided limit exists if and only if both one-sided limits exist and are equal. So "the 2-sidedd limit exists" is exactly the same as "the one sided limits exist and agree". How can the former be irrelevant, but the latter be the key, if they are equivalent? –  Arturo Magidin Nov 26 '11 at 21:33
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@AdamSmith - The content of a question is what's important, not who asked it. If Terry Tao came in and asked a question you deemed below his station, would you downvote it? I cannot imagine any situation in which people should avoid asking a question when they don't know the answer - even if you think it "ought" to be within their ability, all you should worry about is giving a good answer for the benefit of the online math community as a whole. –  Zev Chonoles Nov 29 '11 at 16:01

1 Answer 1

Let $f: \mathbb{R}\to\mathbb{R}$ be a function that is not continuous (or undefined) at $x_0 \in \mathbb{R}$. We can try to remove the discontinuity by replacing $f$ with the "patched" function $g$ given by

$$g(x) = \begin{cases} f(x), &x \neq x_0\\C, &x = x_0\end{cases}$$

for some real number $C$. Almost immediately from the definition of continuity, it follows that $g$ is now continuous at $x_0$ if and only if $$\lim_{x\to x_0^+} f(x) = \lim_{x\to x_0^-} f(x) = C.$$ In other words, the singularity is removable if and only if the left and right limits of $f$ near the singularity exist, and are equal.

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This is exactly the obvious answer that Daniel gave above and you criticized. –  Adam Smith Nov 26 '11 at 21:18
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@Mathemagician1234 : I have no idea what kind of point you are pressing here. It's just the definition of a limit -- nothing subtle is happening here. And I'm not talking about rigorous calculus courses, just ordinary calculus courses that discuss limits (without properly defining them). I don't keep copies of calculus books on hand when I'm not teaching it, so I can't give a precise reference, but this kind of thing is definitely discussed in, e.g., Stewart's book. –  Adam Smith Nov 27 '11 at 4:54
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@Mathemagician1234 : I don't see any material difference between what you describe in Stewart and what is contained in the above answer. –  Adam Smith Nov 27 '11 at 17:36
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@Mathemagician1234: I have removed an irrelevant remark from the above comment. And as I am sure you are aware, each user is only able to vote once on any given post, so it does not make sense to blame a single user for the total score of a post. –  Zev Chonoles Nov 28 '11 at 1:21
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@Zev Alright. I'm trying not to get myself thrown off here. I think I was very civil about it. I'm trying very hard to be. –  Mathemagician1234 Nov 28 '11 at 2:18

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