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Given this function:

Domain = $\{x\in \Bbb R: x\geq 0\}$ $f(0)=0$ and $f(x)=1$ for $x > 0$

The function is discontinued in $x=0$ but what kind of discontinuity?

Calssification: https://en.wikipedia.org/wiki/Classification_of_discontinuities

I think that the point $x=0$ is not any of the discontinity in the classification. Do you confirm that?

asv
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    Yes, but we cannot do the limit on left the zero (the limiti on left the 0 is not defined). I think that at rigour this discontinued at 0 is not in anyone of the classification. – asv Feb 20 '18 at 12:52
  • What's your guess and reasoning for it? – M. Winter Feb 20 '18 at 12:53
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    Intuitively, a discontinuity $x_0$ of a function $f$ should be removable precisley when one only needs to remove the origninal value of $f$ at $x_0$ and replace it by an alternative value to make it continuous at $x_0$. This is the case here. Conversely, if $x_0$ is a jump discontinuoity, one always needs to change the values of $f$ in a neighborhood of $x_0$ in order to make $f$ continous at $x_0$. – H1ghfiv3 Feb 20 '18 at 12:54
  • My guess is that in the calssification (removable or jump discontinuity) is required both limits (right and left) but here we have only right but left limit is not defined. – asv Feb 20 '18 at 12:57
  • @BerniWaterman Yes in intuition, but at rigour the definition required both limits. My intention is exactly see that the rigour definition doesn't capture complety the intuition. – asv Feb 20 '18 at 12:59
  • In order to distinguish between a removable and a jump discontinuity at a point $x_0$, you are right that one always needs $f$ to be defined in a neighborhood of $x_0$, as one always needs both left-and right-handed limits. This is never possible if $x_0$ is the maximum (resp. minimum) of $dom(f)$, yet we can still determine whether or not $f$ is continuous at $x_0$ and can still distinguish between different types of discontinuity behaviour at $x_0$. – H1ghfiv3 Feb 20 '18 at 13:15

2 Answers2

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This is obviously a jump discontinuity if we have $f(x)=0$ for $x\le 0$ since$$osc_f(0)={(\sup-\inf)}_{|x|<\epsilon} f(x)=1-0=1$$but here the domain only contains $x\ge0$ so the discontinuity in this question is removable

Mostafa Ayaz
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Let me make a definition of the different discontinuity types that do not require left and right-sided limits at a given point $x_0$, but are easily seen to coincide with your definition whenever both exist. Thus, let $dom(f) \subseteq \mathbb R$ be a non-empty set, $f: dom(f) \to \mathbb R$ a function and $x_0 \in dom(f)$. Suppose that $f$ is not continuous at $x_0$. Then $x_0$ is

a) a removable discontinuity of $f$, if there exist a function $\hat{f}: dom(f) \to \mathbb R$ with $\hat{f} = f$ on $dom(f) \setminus \{x_0\}$, such that $\hat{f}$ is continuous at $x_0$.

b) a jump discontinuity if is not a removable discontinuity, but there exist a real number $k \in \mathbb R$ with $k \neq 0$, such that the map $g: dom(f) \to \mathbb R$ defined by $$g(x) = \begin{cases} f(x) & x \in (-\infty,x_0] \cap dom(f) \\ f(x) + k & x \in (x_0, \infty) \cap dom(f)\end{cases}$$ has a removable discontinuity at $x_0$.

c) An essential discontinuity if neither a) nor b) hold.

In this terminology, it is clear that $x_0 = 0$ of your original function $f$ is a removable discontinuity, and not a jump discontinuity.

H1ghfiv3
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