Discontinuity of f(x) = {0 in 0 and 1 for x>0} 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?
 A: 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
A: 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.
