A question regarding (ε, δ)-definition of limit Definition

$\lim_{x \rightarrow a} f(x) = L$ means, $\forall \epsilon >0, \exists \delta >0; 0 < |x-a|< \delta \Rightarrow |f(x) - L|< \epsilon$, with $x, a, L \in \mathbb{R}$.

I understand this is THE definition of limit and I should follow as it is stated
But I just can't understand why it is written this way.
To me,

$\lim_{x \rightarrow a} f(x) = L$ means, $\forall \delta >0, \exists \epsilon >0; 0 < |x-a|< \delta \Rightarrow |f(x) - L|< \epsilon$, with $x, a, L \in \mathbb{R}$

this way of writing(choosing $ \delta > 0$ first and finding corresponding $\epsilon$) seems to work as well as choosing epsilon first.
For every function I(first year in college) can think of, I can pick any delta I want FIRST and still find corresponding $\epsilon$. Rest of the definition also fits
So here's my question:
Is it possible that we choose delta first and still make it work?
 A: Let $f(x)$ be a function that hops crazily between $0$ and $1$, say $0$ at rationals and $1$ at irrationals. For any $\delta\gt 0$, there is an $\epsilon$, namely $17$, such that whenever $|x-a|\lt \delta$, we have $|f(x)-f(a)|\lt \epsilon$.
A: We need to understand the concept of limit informally. The concept is introduced to study the behavior of a function $f(x)$ when the values of $x$ are near a certain specific value say $x = a$. Our goal is to see if the values of $f(x)$ also lie near some specific value $L$ when $x$ is near $a$. This "near to" has to be as good as it is possible and hence we need to quantify them using arbitrary positive numbers like $\epsilon$.
In the definition of limit $\lim_{x \to a}f(x) = L$ we want to ensure that $f(x)$ can be made arbitrary near to $x$ by making $x$ sufficiently near to $a$. Thus the difference $|f(x) - L| $ has be made less than any arbitrary $\epsilon > 0$ by making $0 < |x - a| < \delta$ for some suitably chosen $\delta$ depending upon $\epsilon$. According to your version when $\delta > 0$ is arbitrary and $\epsilon > 0$ is chosen based on $\delta$ we are not guaranteeing that the difference $|f(x) - L|$ can be made as small as we would like it to be. In fact in your version we just can choose any big number $\epsilon$ without any problem and yet satisfy the inequalities you mention (unless the function is bounded). For example if $f(x) = \sin x, x = 0$ then given any $\delta > 0$ we have $0 < |x - a| < \delta$ always imply $|f(x) - L| \leq |\sin x| + |L| \leq 1 + |L| < \epsilon$ if we choose $\epsilon $ greater than $|L| + 1$. And this happens for any number $L$.
So your definition does not try to specify any property of a specific $L$.
