# 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?

• The Wikipedia article is misleading. It just uses epsilon and delta as variables without explaining what those variables really are. At the start of the article drill down on epsilon and delta and see what they really are. – MaxW Sep 13 '16 at 14:55
• @MaxW The good thing with Wikipedia is that you can edit it. So if you have a better explanation, feel free to improve the article. – Najib Idrissi Sep 13 '16 at 14:58
• If you want the constant function to satisiy your statment you have to replace $\lt$ by $\le$. – miracle173 Sep 13 '16 at 14:59
• The terms epsilon and delta have links. You don't want to write definitions in 500 places since any change in the definition leads to having to edit 500 articles not one. – MaxW Sep 13 '16 at 15:14
• Proofs and examples are not the same thing. – Nic Hartley Sep 13 '16 at 18:57

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.

• Thnx a lot, I was eagerly waiting for a counter example, and that's a great one – codetalker Sep 13 '16 at 15:06
• +1 for the last paragraph which gives the crux of this question. Too good! – Paramanand Singh Sep 13 '16 at 16:40

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.

• Thnx for the explanation, I guess I will have to change my intuition then – codetalker Sep 13 '16 at 15:05
• @Siddhant Your intuition is probably fine, but you need to learn to formalize it better. – Najib Idrissi Sep 14 '16 at 6:07

A more severe problem with this definition is that the $\iff$ in the inequality implies that $f$ takes level sets of distance to $a$, to level sets of $f$, so that (on intervals) the only continuous functions according to this concept of limit are linear.

To formalize the property "$x$ closer to $a$ implies $f(x)$ closer to $L$", the arrow in the definition should be only in the forward direction, $\implies$. Although that is a stronger property than existence of a limit, it is also true that this property holds in most cases where there is a limit.

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).

It is a definition. You can accept it as it is, and prove theorems based on that definition. But it is not the usual definition of a limit, and has some severe disadvantages.

One: f (x) = 0 has no limit :-)

Two: A function can have more than one limit at a minimum or maximum. Take f (x) = $x^2$. Any l ≤ 0 is the limit at a = 0 with this definition.

Three: A function must be (kind of) symmetric to have a limit - we typically need f (a +/- eps) = f (a) +/- delta.

Four: The definition doesn't use just local properties. sin (x) has no limit at a = 0 because sin (π) = 0 again.

I haven't thought about it too much, but if f (x) has a limit at two points a and b, that might restrict substantially what f (x) can look like.