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 $x = 0$.
Best Answer
Yes
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