TITLE: Can this be considered a "rigorous" definition of a limit?
QUESTION [12 upvotes]: I would like to show my definition of a limit, along with an example:

Let $f:D\to\mathbb{R}$, where $D\subset\mathbb{R}$. Let
  $a\in\mathbb{R}$. If some $l\in\mathbb{R}$ exists, such that for all 
  $x,y\in D$, $|a-x|<|a-y|\iff |l-f(x)|<|l-f(y)|$, then $l$ is called
  the limit of the function $f$, and subsequently: $$\lim_{x\to
 a}f(x)=l$$

Using this to prove a standard limit(example is taken from this page):
We wish to prove the following limit:
$$\lim_{x\to5}(3x-3)=12$$
According to my definition, here $a=5$, and $l=12$.
Now, let $f(x)=3x-3$, and $domain(f)=D$. Let $x,y\in D$ such that the following inequality holds:$$|5-x|<|5-y|$$
Now, following steps are followed:
$$ |5-x|<|5-y| $$
$$\implies 3|5-x|<3|5-y| $$
$$\implies |3(5-x)|<|3(5-y)|$$
$$\implies |15-3x|<|15-3y| $$
$$\implies |15-3x|<|15-3y| $$
$$\implies |12-f(x)|<|12-f(y)|$$
and thus, we proved the statement.  
Is this definition correct?

REPLY [2 votes]: According to your definition, $\lim_{x\to1}x^2 \neq 1$, because, $|0-1|<|(-1)-1|$ and yet $|f(0)-f(1)| > |f(-1)-f(1)|$.
Thought of another way: Consider $x^2$. As $x$ goes from -1 to 1, the value of $f(x)$ first moves away from 1, and only then starts getting closer. With the usual definition, we only care about an arbitrarily small neighborhood of the point - So it only matters that it "eventually" gets closer. But your definition is global, and the behavior of the function away from the point ruins the limit.
This is somewhat similar to the example $x\sin(1/x)$, but it shows that you don't need crazy infinitely oscillating functions to see the flaws in your method. Of course, you can remedy the global dependence by allowing the inequality to hold just in a neighborhood, but that further complicates the definition.
Also, even if your function captures some sense of "global limit", in order to call it "the limit" and give it a notation, you have to prove that a number satisfying the requirement is unique. (It is not, as shown in other answers).