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I learned that if a function is undefined at a point and becomes discontinuous because of that, it is actually a removable singularity, whereas if we set it to be different valued through a piecewise function, it is a removable discontinuity. So by my logic, the function above has a removable singularity at x=2. Am I correct?

Nublets
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The function $f(x) = \frac{x-2}{x-2}$ (on the implied domain $\Bbb R\setminus \{2\}$) has a removable singularity at $x = 2$. It does not have a discontinuity there (removable or otherwise) because it is a continuous function. For a function to be discontinuous, there must be a discontinuity and such a discontinuity must lie in the domain of the function, by definition. The point $x = 2$ is not in the domain, and therefore the function cannot possibly be discontinuous there.

Arthur
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  • +1. A related discussion: https://math.stackexchange.com/questions/1087623/is-function-f-mathbb-c-0-rightarrow-mathbb-c-prescribed-by-z-rightarrow – Hans Lundmark Sep 25 '19 at 13:06