What type of discontinuity is found in this graph?
$$ f(x) = \begin{cases} \dfrac{1}{x} && \text{when $x > 0$}\\ 4 && \text{when $x < 0$} \end{cases} $$
What type of discontinuity is present when $f(0)$ ?
I'm currently taking calculus and from what I've learnt, there are three kinds of discontinuities. Point/removable discontinuity, infinite discontinuity and jump discontinuity. I know definitely that the graph above does not have a point discontinuity. However, I'm unsure of whether the graph above contains an infinite discontinuity or jump discontinuity. Can someone provide me an answer with an explanation as to why?
To generalize this question, the situation could be: As $x$ approaches $a$ from one side, there is a finite limit. As $x$ approaches $a$ from the other side, $y$ approaches positive or negative infinity. However, $f(a)$ is non-existent. What type of discontinuity can be found in $f(a)$?
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$\begingroup$Jump discontinuity occurs when sided limits exist. It means that
$$\lim_{x\to a^+}f(a)\in\mathbb{R} \quad\text{and}\quad \lim_{x\to a^-}f(a)\in\mathbb{R}$$
So clearly this is not a jump discontinuity.
The "infinite discontinuity" is often called "essential discontinuity" so that it's not misleading. This discontinuity is essential as the limit from the right doesn't exist (see the Wikipedia).
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