7
$\begingroup$

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?

$\endgroup$
2
  • 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
3
$\begingroup$

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 $\lim\limits_{x\to a}f(x)=L$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
-1
$\begingroup$

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)).

$\endgroup$
1
  • $\begingroup$ 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

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.