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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?
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.
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)).
