Can this be considered a "rigorous" definition of a limit?

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?

Solution 1:

No, this is not equivalent to the usual definition of a limit. Consider for example $$f(x) = \begin{cases} x \sin(1/x) & x \neq 0 \\ 0 & x = 0 \end{cases}$$

Then $\lim_{x \to 0} f(x) = 0$ as can be easily checked (note that $|f(x)| \le |x|$ for all $x \in \mathbb{R}$).

However with your definition then $f$ wouldn't have this limit. Indeed, applied to $a = 0$ and $l = 0$, this would mean that we would have $$|x| < |y| \iff |x \sin(1/x)| < |y \sin(1/y)|.$$ This is false. Let $x = \frac{2}{3\pi}$ and $y = \frac{1}{\pi}$. Then $|x| < |y|$, however $|f(x)| = \frac{2}{3\pi} > |f(y)| = 0$.

I guess what you were going for was "the closer $x$ is to $a$, then the closer $f(x)$ is to $l$"? The problem with that, as you can see, is that $f(x)$ can oscillate a lot around $l$ even though it gets closer and closer to it.

Solution 2:

This definition says "whenever $x$ gets closer to $a$, $f(x)$ gets closer to $l$". But this is not the usual definition of limit. For example, let's consider $f(x)=x^2$, and $a=0$. Then, since $$\lim_{x \to a} f(x)=0^2=0$$ it is obvious that whenever $x$ gets closer to $0$, then $f(x)$ gets closer to $0$. However, it gets closer to every other negative number as well.

So, according to your definition of limit, it would be that every negative number $l <0$ satisfies $$\lim_{x \to 0}x^2=l$$ which is not the usual notion of limit.