Talk:Classification of discontinuities

message: This article was confusing continuous functions (which are continuous on their domains) with continuous funciton on the real numbers.  So technically all the example functions are continuous if they are not defined at x=1.  I have edited the examples so that they are all total functions, removing this confusion.  I would like to make the jump discontinuity half-continuous but then I would have to edit the image. —The preceding unsigned comment was added by 2006 80.57.33.218 (talk • contribs) .
: You are right of course. I did not think of that when i added the examples.  Oleg Alexandrov (talk)
: I would like to add that a jump discontinuity can be defined as a place where the left and right limits both separately exist, but are not equal.  This is quite different from an 'essential' singularity, as the function Exp(-1/x^2) certainly has an essential singularity at the origin when thought of as a complex function, but I would argue that it has a removable discontinuity when thought of as a function of real x only.  It is certainly discontinuous at x = 0, simply because it is not defined there.  We cannot say that it is continuous there, so it's not continuous (discontinuous) there.  I understand where the other editors are coming from when they say that discontinuity can only be defined at a point in the domain of the original function, but I think that this is splitting hairs a bit too much and can be confusing to the uninitiated reader.  It is incorrect to classify a point at which the function is not defined, but for which the limit exists, as a 'removable singularity', as this term means that the limit exists from all COMPLEX directions.  The function f(x) = \sqrt(x) for x > 0 and 0 for x < 0 is certainly undefined at zero, and the limit certainly exists (when thought of for only real x), but defining f(0) = 0 will not remove the singularity at x = 0.  The function is still 'singular', but it will now be continuous.  This is why I prefer the term 'removable discontinuity'.  Any comments on this take?  (from Jonathan Osborne)— Preceding unsigned comment added by Physicself (talk • contribs) 01:08, 4 May 2012 (UTC)
thread_title: Confusion

message: I agree with the above criticism of Stewart's text. Defining discontinuity of f at x for x not in the domain of f seem to indicate a lack of coherent language that allows to express properties of f without confusion. Rather than saying that function f, undefined at x, has a non-removable discontinuity at x, it would be more appropriate, and consistent with set theory, to say that there is no continuous extension of f over x. The latter means that whatever value will be assigned to x, the extended function will not be continuous at x.
To also cover the case when x is in the domain Dm(f) of f, one can say that:
"function f restricted to Dm(f)\{x} has no continuous extension over x."
This carries all the information of the phrase "f has a non-removable discontinuity at x" without confusion when f is a continuous function.
Removable discontinuity may be similarly replaced with the existence of a continuous extension.
I can speculate that someone (a colleague of mine suggested Richard Courant as the inventor of the above taxonomy of discontinuous at x with no regard whether x is in Dm(f) or not) interpreted "whatever value" as "not defined value", and then decided to include "undefined value" just to make it more general.  — Preceding unsigned comment added by 76.171.57.52 (talk) 06:55, 29 May 2015 (UTC)
thread_title: A constructive suggestion

message: The current article has examples of discontinuities, but no definition of the general phenomenon. Do we obtain the definition of "x = a is discontinuity of f(x)" by negating the definition of "f(x) is continuous at x = a"?  If so, that would be a start in settling the controversies that arise about values x =a that are not in the domain of a function f(x).   
We could analyze whether the definition of "f(x) is continuous at x = a" can only be satisfied by values of x in the domain of f.  That would make any value not in the domain of f(x) automatically a discontinuity of f(x).  However, I don't know if that conclusion would be harmonious with the assertion that discontinuities must fall into one of the three categories illustrated in the current article.  For example, if we define  f(x) = \sqrt{ x(x-2)}  to have a domain consisting of only those real numbers where of  \sqrt{(x)(x-2)}  is a real number then which type of discontinuity would  x = -1.3  be?
There is also the hair-splitting question of whether a discontinuity of a function of a real variable must be a real number.  For example, my brother's red socks are not in the domain of a function of a real variable, but presumably my brother's red socks are not a discontinuity of such a function.  Perhaps (in the case of real valued functions of a a real variable) the technicality is that "f(x) is continuous at x = a" is only defined for real numbers "a", so its negation is likewise only defined for real numbers "a".  
Tashiro~enwiki (talk) 02:53, 1 March 2017 (UTC)
replies: text: This can certainly not be defined here nor elsewhere. A function can only be continuous or discontinuous at points in its domain. Otherwise e.g. the real log-function would be discontinuous for all negative variable-values or any real function would be discontinuous at all non-real complex points. --Felix Tritschler (talk) 00:26, 15 April 2024 (UTC)
thread_title:  Can we define "x = a is a discontinuity of f(x)" by negating "f(x) is continuous at x = a" ?

message: In the first paragraph of This article, it says "This article describes the classification of discontinuities in the simplest case of functions of a single real variable taking real values." Isn't that redundant? Alej27 (talk) 23:00, 18 November 2020 (UTC)
thread_title:  Isn't the phrase "real variable taking real values" redundant?