How to make sense out of the $\epsilon-\delta$ definition of a limit?

The definition can be read, in human words:

For every positive $\epsilon$, there exists such a positive $\delta$ that if $|x-c|$ is smaller than $\delta$, but larger than $0$, then $|f(x) - L|$ is smaller than $\epsilon$.

First of all, lets's get the double inequality out of the way. Basically, $0<|x-c|$ is just saying that $|x-c|$ is not equal to $0$ (since it can't be negative), and saying that is just saying that $x-c$ cannot be $0$, or in other words, that $x$ is not allowed to equal $c$.

Now to a non-mathematician, that still makes very little sense, but take into account that $|a-b|$ is really the distance between numbers $a$ and $b$.

So, we can translate the definition into

For every positive $\epsilon$, there exists such a positive $\delta$ that if $x$ and $c$ are two distinct numbers and the distance between them is smaller than $\delta$, then the distance between $f(x)$ and $L$ is smaller than $\epsilon$.

But that still does not ring quite "natural" But what does the "for all $\epsilon$, there exists a $\delta$" in the beginning mean anyway? Well it means that whatever $\epsilon$ you give me, I can find a $\delta$ such that the condition will be true. So:

No matter what $\epsilon$ you choose, I can find such a positive $\delta$ that whenever you take any $x$ near (but not equal to) $c$ that is less than $\delta$ away from $c$ which are closer together than $\delta$, $f(x)$ will be closer than $\epsilon$ away from $L$.

Getting warmer to something readable? Well, let's get rid of the variables even further:

No matter how close you want $f(x)$ to be to $L$, I can tell you how close to $c$ you need to pick your $x$, and if you pick your $x$ that closely, then I can guarantee that $f(x)$ will be as close to $L$ as you originally wanted it to be.

This is very similar to what MPW wrote in comments: