How does the $\epsilon - \delta$ actually ensure that $L$ is the limit? 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. 
 A: 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).
A: By limit, let say we mean that the closer I am to $a$ then the closer I am to $L$. 
The understanding should be that I could get as close as I want to $L$ and that I at some point a don't go too far away from $L$. 
To understand the definition, you might want to take the wiewpoint of the limit and not the viewpoint of the variable. 
Assume you have ordered a cab and you can see on your famous app that the cab is moving towards you. Because streets might not be perfectly regular, the cab do some detour, however it does come closer and closer.
So you hope that if you wait enough time he will get closer that 100m, closer than 10m, clsoer that 1m,...  
A: Well, one approach would be to consider some other point, $R$, and show that $f$ has to eventually always be closer to $L$ than to $R$. 
Assume that $\displaystyle\lim_{x\to a} f(x) = L$, and let $R$ be any point other than $L$. Now, take any $\varepsilon$ that satisfies $0<\varepsilon<\frac{|L-R|}{2}$. This ensures that 

$\text{if } |f(x)-L|<\varepsilon \text{, then } |f(x)-L|<|f(x)-R|.$

By the definition of the limit, there exists some $\delta>0$ where if $|x-a|<\delta$, that is, if $x$ is close enough to $a$, then $f(x)$ must always be closer to $L$ than to $R$. 
But we can do this for every point $R$ that isn’t $L$! We can always force $f(x)$ to be closer to $L$ than to every single other point, so $f(x)$ must go to $L$. 
A: I think the following illustration might clarify the limit concept for a one variable function.

So as $x$ gets closer to $a$, $f(x)$ gets closer to $L$
