TITLE: Does |x-c| and $\delta$ refer to one same thing: the distance between c and x in the x-axis, well why |x-c|<$\delta$?
QUESTION [1 upvotes]: The way i understand $\epsilon$ and $\delta$ proof is this:
Given a  $\lim_{x\to c}f(x) = L $ suppose we have a $\epsilon > 0$ (a change for L over the y-axis) for every $\epsilon$>0 there gonna be a $\delta$>0 (again a change for c in the x-axis), so this is it $\delta$ and $\epsilon$ is small change of the values L and c.
But the definition -$\delta$<x-c<$\delta$  <=> |x-c|<$\delta$, says something else i think, it says the change over the x-axis is the distance |x-c| which means $\delta$ is definitely not the change , because if it's the case it would be |x-c| = $\delta$, so what is exactly is delta,  this expression-$\delta$<x-c<$\delta$ is saying for me that $\delta$ is the interval in which the change is happening , can you clarify please.

REPLY [3 votes]: $\delta$ and $\epsilon$ both represent ranges of change in the appropriate axis. We don’t get $|f(x)-L|=\epsilon,$ either, we get $|f(x)-L|<\epsilon.$
So we are saying that we can restrict the range of $f(x)$ to $L-\epsilon<f(x)<L+\epsilon$ if we restrict the range of $x$ to $c-\delta<x<c+\delta$ and $x\neq c.$
It turns out the $\epsilon-\delta$ definition is equivalent to:

Equivalent definition of $\lim_{x\to c} f(x)=L:$


For any real $y_1,y_2$ with $y_1<L<y_2$ there exists a pair of reals $x_1,x_2$ With $x_1<c<x_2$ such that for $x_1<x<x_2$ (and $x\neq c$), we have $y_1<f(x)<y_2.$

This is saying, “For any range around $L,$ there is a range around $c$ where, if $x$ is in the range around $c,$ but $x\neq c,$ then $f(x)$ is in the range around $L.$”
$\epsilon-\delta$ is useful because “all ranges” is a bit much to handle. $\epsilon-\delta$ becomes something we can handle computationally.
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