For example the graph $f(x) = 1/x$ approaches $\infty$ at $x=0$ but we would not say this is an infinite discontinuity, just an asymptote, correct? Unless we specifically said the domain of the function included $x=0$? Or must domains by definition exclude discontinuities and undefined points?
Do discontinuities only exist if we can "split" the domain into two new non-empty intervals that each exclude the discontinuity?
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$\begingroup$Continuity is only defined at points within the domain of the function . Since a discontinuity is usually defined as a point where the function is not continuous, that must be within the domain as well. It would be meaningless to say that $\,\sqrt{x}\,$ is discontinuous at $\,x=-1\,$, for example.
Or must domains by definition exclude discontinuities and undefined points?
The domain can not include "undefined points", by definition.
The domain can include discontinuity points, and often does, for example the integer part function $\,f(x) = \lfloor x \rfloor\,$ is defined on $\,\mathbb{R}\,$, which includes all the discontinuity points $\,x \in \mathbb{Z}\,$.
Consider $f:\mathbb R\to \mathbb R$ where $f(x)=\begin{cases}1, \text{ if } x\in \mathbb Q\\ 0 ,\text{ if } x\notin \mathbb Q \end{cases} $. This function is discontinuous everywhere.
But, in your example, $f(x)=\frac1x$ isn't defined at $x=0$. You could define $g$ to be the same as $f(x) $ away from zero, and give it some value, say $1$, at $0$. Then $g$ would not be continuous at $0$...
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