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$\exists \epsilon>0$ $\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.

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You can make it look more like continuity by changing the direction of the inequality involving $\delta$. But it makes no difference here, since whatever $x\neq x_0$ you want, you can always find an appropriate $\delta$ either way. – Marc van Leeuwen Nov 28 '11 at 14:36
up vote 3 down vote accepted

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.

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