# Is it okay to say that a function that's non-continuous at a point is continuous on its domain?

I was watching a video on the main continuity theorem, and the following slide came up:

Even though $$f(x) = \sin \frac{1}{x}$$ is not continuous at $$x = 0$$, we could still say it is continuous, or continuous on its domain, correct? Since its domain doesn't include $$x = 0$$.

An example illustrating this for an isolated point is $$f(x)=\sqrt{x^4-x^2}$$ which has domain $$(-\infty,-1] \cup \{0\} \cup [1, \infty)$$.
This $$f(x)$$ is continuous on its domain according to the definition of continuity, even at the isolated point $$x=0$$, as there are no other points near enough to $$0$$ to demonstrate discontinuity
Yes. Since the assertion “$$f$$ is continuous” means “for each $$a$$ in the domain of $$f$$, $$f$$ is continuous at $$a$$”.
Yes, that's right. $$f(x)$$ is continuous on every point in its domain; there are no discontinuities where the function is defined.