Quickly writing out the definition,
$\lim\limits_{x \to a} f(x) = L$ iff $ \forall \epsilon >0, \exists \delta >0, \ s.t. \ 0<|x-a|<\delta \implies |f(x)-L|< \epsilon$
Here is what I think it means,
For any small distance $\epsilon >0$, I have to find at least one deleted $\delta$ nbd of $a$ for which the distance between $f(x)$ and $L$ is even less than $\epsilon$ if every $x$ is taken from the deleted nbd of $a$.
This is just translating that expression into simple English. The thing is that, I "understand" the "expression" but I just can't see why that would imply or tell us that $L$ is indeed the limit of the function $f(x)$ at $x=a$
Again, please note that, I don't need an explanation for the mathematical expression, I need help understanding why would that imply $L$ is the limit. What is the logic here?
Edit: I understand limit to be a particular specific value that the function approaches when $x$ approaches $a$. It does not assert that $f(x)$ is $L$ at $x=a$ i.e $f(a)= L$. $f(a)$ might as well not be defined for all we care.
Best Answer
Let's say you fix some $\varepsilon > 0$ and prove that there is $\delta$ such that $|f(x) - L|<\varepsilon$. That can be rewritten in the form $f(x)\in (L-\varepsilon, L + \varepsilon)$ for $x$ sufficiently close to $a$. This tells you that if the limit at $a$ exists, it should be in the interval $[L-\varepsilon, L + \varepsilon]$. Now, if you prove the same thing for every $\varepsilon>0$, it tells you that the limit at $a$ should be in
$$\bigcap_{\varepsilon>0}[L-\varepsilon, L + \varepsilon] = \{L\}.$$
So we say $\lim_{x\to a} f(x) = L$.
Note that this intuition is basically squeeze theorem (and that's why I use segment instead of open interval).