Nonremovable Discontinuity

The repair test is direct. To make $f$ continuous at $x = c$ you would define $f(c)$ to be $\lim_{x \to c} f(x)$, so that limit has to exist first. When the one sided limits disagree you have a jump, when either one runs off to infinity you have an infinite discontinuity, and when the values keep oscillating, as $\sin(1/x)$ does at $0$, there is no value to assign. None of the three can be repaired. Some texts reserve the name essential discontinuity for the narrower case where a one sided limit itself fails to exist, infinite or oscillating, which excludes jumps.

Compare the case that does work. For $f(x) = \frac{x^2 - 4}{x - 2}$ the limit at $x = 2$ is $4$, so defining $f(2) = 4$ closes the hole and the repaired function is continuous. A vertical asymptote offers you nothing to define: no single number sits at the end of an output that grows without bound.

The two sided test assumes there are two sides. Continuity at an endpoint of the domain is one sided, so for $f$ on $[0, \infty)$ with $f(0) = 5$ and $f(x) = x$ for $x > 0$, the two sided limit at $0$ does not exist and the break is still removable: resetting the single value to $f(0) = 0$ makes $f$ continuous on its domain. Apply the criterion at interior points, and the matching one sided limit at the ends.