Tukukkk

This is a question regarding the definition of continuity.

My understanding of continuity is that a function is continuous at a point when it holds that $$lim_{xto a^-}f(x) = f(a) = lim_{xto a^+}f(x) quad quad (1)$$

The book I'm currently reading has this image:

Note here that $f(x)$ is defined for $x=3$, but $g(x)$ is not.

This is followed by text stating that

g(x) is continuous because $D_g = [0, 6]text{\}{3}$, thus it is continuous for all values in its domain.

My point of contention here is that, how can we say that it is continuous at $x=3$ when $g(3)$ does not exist? Referring to the aforementioned definition $(1)$ that the limits converge to the actual value at this point.

I would have immediately declared both cases as jump discontinuities.

Am I mistaken here? Does $g(x)$ illustrate an exception to $(1)$?

$G$ is continuous on the domain $[0,3)cup(3,6]$.

Referring to the aforementioned definition (1) that the limits converge to the actual value at this point.

3 is not in the domain. For every point in the domain of $g$, we have the required convergence.

The function is continuous everywhere in the interval except that point deleted from the domain, it's more a nuance of the language than anything else. Choose any point that is not $3$ in that interval: you can then find left- and right-hand limits to that point and show they're equal.

Although not mathematically rigorous, a good rule of thumb is if you can draw the graph without lifting your pencil, it is continuous.