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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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    I don't understand why this question got 6 downvotes. I would not upvote it but now I will to compensate for inappropriate downvotes. +12012-10-12

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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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    @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.2011-11-28