0
$\begingroup$

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

enter image description here

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 $x = 0$.

$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Yes. Since the assertion “$f$ is continuous” means “for each $a$ in the domain of $f$, $f$ is continuous at $a$”.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Yes, that's right. $f(x)$ is continuous on every point in its domain; there are no discontinuities where the function is defined.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

If you define a function to be continuous iff it is continuous at every point in its domain, then yes. However, this can cause confusion. I think it's best to always specify where a function is continuous, whether it be the entire domain or some subset of it.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.