TITLE: Wrong Definition of a Limit
QUESTION [7 upvotes]: 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?

REPLY [3 votes]: 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 $\lim\limits_{x\to a}f(x)=L$.

REPLY [2 votes]: 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.