Wrong Definition of a Limit What would this mean:


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


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 $0 &lt; |x − a| &lt; \delta$, we will have $|f(x) − L| &lt; \epsilon$.

Does this show that there simply exists an interval where $f(x)$ is a constant function?
Definitely it would mean $f$ is constant, and equal to $L$, on a neighborhood of radius $\delta$ about $a$.  If you write $\forall\varepsilon&gt;0\ \exists\delta&gt;0$ rather than the other way around, then it is a weaker assertion: that $\lim\limits_{x\to a}f(x)=L$.
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)).