What would this mean:

$\exists \delta >0$ such that $\forall \epsilon > 0$ and $\forall x$ satisfying $0 < |x-a| < \delta$, then $|f(x)-L| < \epsilon$

I am pretty confused by the symbols too...

Here's I read it:

There exists a delta larger than zero such that for any epsilon larger than zero and for any $x$ satisfying $0 < |x − a| < \delta$, we will have $|f(x) − L| < \epsilon$.

Does this show that there simply exists an interval where $f(x)$ is a constant function?

  • 3
    $\begingroup$ Yes....it does. $\endgroup$ – Bill Cook Oct 18 '11 at 23:57
  • $\begingroup$ Thanks for the edit Sivaram, I hope this helps people who look at this in the future! $\endgroup$ – MathMathCookie Oct 19 '11 at 1:05

Definitely it would mean $f$ is constant, and equal to $L$, on a neighborhood of radius $\delta$ about $a$. If you write $\forall\varepsilon>0\ \exists\delta>0$ rather than the other way around, then it is a weaker assertion: that $\lim\limits_{x\to a}f(x)=L$.


Yes. Specifically it implies $f(x)=L$ on some interval $(a-\delta,a+\delta)$. This is because the statement says that $f(x)$ is arbitrarily close to $L$ for $x$ within this interval (since we can choose $\epsilon>0$ as small as we desire), and in the real numbers arbitrarily close means equality.

  • $\begingroup$ I'm sorry I couldn't pick your answer - his came first :( Thank you so much though! $\endgroup$ – MathMathCookie Oct 19 '11 at 1:06
  • $\begingroup$ @anon How can arbitrarily close imply equality, especially given the fact that $\varepsilon > 0$ ? $\endgroup$ – curryage Mar 28 '14 at 11:09
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    $\begingroup$ @curryage Isn't it extremely obvious? Can you think of two real numbers that are arbitrarily close but distinct? Do you know what "arbitrarily close" means? $\endgroup$ – anon Apr 1 '14 at 7:37
  • $\begingroup$ @anon Ok. Makes sense. $\endgroup$ – curryage Apr 1 '14 at 10:17

Not really. f(x) is still the function itself - it does not become a constant function.

What it says is that if you select an appropriate δ (often small enough), you can get the distance to f(x) smaller than any ε. Very often δ is built on ε, been restricted to the f:=x relationship, the difference between f(x) and L is always smaller than ε. Now L is the limit, meaning you are close enough to f(x), but you are not at the point of (x, f(x)).

  • $\begingroup$ Note that OP has written the usual limit definition with the $\forall\epsilon>0$ and $\exists\delta>0$ backwards. $\endgroup$ – anon Oct 19 '11 at 0:07

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