$\exists \epsilon>0$ $\forall\delta>0: |x-x_0|> \delta \to $ $|f(x) - f(x_0)| < \epsilon$
It is very similar to the continuity of the function at a point, but it is not it. I hope for your help!
P.S. Sorry for my bad English.
$\exists \epsilon>0$ $\forall\delta>0: |x-x_0|> \delta \to $ $|f(x) - f(x_0)| < \epsilon$
It is very similar to the continuity of the function at a point, but it is not it. I hope for your help!
P.S. Sorry for my bad English.
The condition is that $f$ is bounded.
Since $\varepsilon$ does not depend on $\delta$, we might as well take $\delta\to 0$ and get that
$\exists \varepsilon > 0\ \forall x, x_0\ |f(x) - f(x_0)| <\epsilon$ which is true precisely when $f$ is bounded above and below.