Why is the condition $0<|x-a|<\delta$ necessary in the $\varepsilon$-$\delta$ definition of a limit?

Question

The $\varepsilon$-$\delta$ definition of a limit is as follows:

$$\lim_{x\to a}f(x)=L\iff\forall\varepsilon>0\ \exists\delta>0;\big(0<|x-a|<\delta\implies|f(x)-L|<\varepsilon\big)$$

I've observed that $0<|x-a|<\delta$ is always true, as there always exists a $\delta$ such that $\delta=|x-a|+1$, making the statement $0<|x-a|<\delta$ true (Or even we can choose a very, very large $\delta$ that does not depend on $x$, such as $\delta=10^{10^{1000}}$):

$$0<|x-a|<|x-a|+1$$

This leads me to believe that the first part of the definition is always true. Given this, I've proposed rewriting the $\varepsilon$-$\delta$ definition as follows:

$$\lim_{x\to a}f(x)=L\iff\forall\varepsilon>0;\big(|f(x)-L|<\varepsilon\big)$$

However, upon reflection, I've realized that this rewritten definition seems awkward and incorrect. Despite the first part of the definition appearing to be always true, it's still necessary. The question then arises: what exactly is the issue?

I think that the key lies in understanding the role of $\delta$ in the definition. The statement $0<|x-a|<\delta$ isn't just stating that $x$ is within some interval around $a$. Rather, it's saying that we can make $x$ as close to $a$ as we want by making $\delta$ sufficiently small. The choice of $\boldsymbol\delta$ depends on $\boldsymbol\varepsilon$ (Is this right?), and for each $\varepsilon$, we need to find a corresponding $\delta$ such that if $x$ is within $\delta$ of $a$, then $f(x)$ is within $\varepsilon$ of $L$. This is the essence of the limit concept. So, the first part of the definition isn't just obvious and always true; it's a crucial part of capturing the idea of closeness inherent in the concept of a limit. Am I right?

Answer 1

"capturing the idea of closeness inherent in the concept of a limit" :

This is exactly the point, which can be formalized in modern infinitesimal analysis as follows: whenever $\alpha$ is a nonzero infinitesimal (i.e., infinitely close to zero), $f(x+\alpha)-f(x)$ is also an infinitesimal.

In fact, Cauchy defined continuity of $f$ in 1821 by requiring that an infinitesimal $\alpha$ always produces an infinitesimal change in the function: $f(x+\alpha)-f(x)$.

The $\epsilon,\delta$ formulation (of both limit and continuity) is a long-winded paraphrase of Cauchy's original definition in terms of infinitesimals. In such a definition, as you correctly point out, the $\delta$ must depend on $\epsilon$: if you consider for example the function $f(x)=1000x$, you will see that one can indeed make $f(x+\delta)$ at will $\epsilon$-close to $f(x)$, but one must choose $\delta$ much smaller than $\epsilon$, namely $\delta<\frac1{1000}\epsilon$.

Answer 2

EDIT: In response to your edit, yes, your last explanation is right--the concept of "closeness" is the whole point of the limit definition. If $x$ is "close" to $a$, then $f(x)$ is "close" to $L$.

I am leaving the previous response below in hopes it may help future people.

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So the first section is not always true. What the definition says is if the first part is true, then it implies the second part must be true. In other words, suppose $a \in \mathbb{R}$ is some given number. I like parsing these math notation derivatives as a series of steps, like a game.

Then if $\lim_{x \to a} f(x) = L$, that's a shorthand for saying:

1. For any $\epsilon > 0$, (i.e., you give me some $\epsilon$)
2. There is some $\delta > 0$ such that (i.e., I will give you a $\delta$ based on $\epsilon$ and $a$)
3. For every $x \in \mathbb{R}$, if $|x - a| < \delta$ (i.e., you choose an $x$ such that $|x - a| < \delta$)
4. That implies $|f(x) - L| < \epsilon$ (i.e., I will show you that $|f(x) - L| < \epsilon$)

A few things to note here:

1. Notice $\delta$ is "chosen" in the definition before $x$ is chosen. In fact, $\delta$ can be based on $\epsilon$ and be based on $a$, but not based on $x$, since $\delta$ is chosen first! So that means yours $\delta = |x - a| + 1$ is not even really valid.

2. You have to show that that $\delta$ actually works. That is, you need to say $\delta = [\text{something or other}]$ such that FOR ANY $x$ where $|x - a| < \delta$, we can prove $|f(x) - L| < \epsilon$.

3. The first part of the if-then statement is not always true. However, we don't care in the case when it's not true! An if-then statement tells you "when the first thing is true, then we know it implies the second thing is true." It does not say anything when the first thing is not true.

Hope this helps.

Answer 3

The condition $0<|x-a|<\delta$ is indeed crucial to the definition, and it's not just trivially true.

The $\varepsilon$-$\delta$ definition of a limit is a formalization of the intuitive notion of a limit, which is about how the values of a function $f(x)$ behave as $x$ gets arbitrarily close to a certain value $a$. The definition captures this "closeness" in a precise mathematical way.

The condition $0<|x-a|<\delta$ is saying that $x$ is within a $\delta$-neighborhood of $a$, but not equal to $a$. This is important because we're interested in the behavior of $f(x)$ as $x$ approaches $a$, not necessarily the value of $f(x)$ at $a$ itself.

The choice of $\delta$ indeed depends on $\varepsilon$. For each $\varepsilon>0$, we need to find a corresponding $\delta>0$ such that if $x$ is within $\delta$ of $a$ (but not equal to $a$), then $f(x)$ is within $\varepsilon$ of $L$. This is the crux of the limit concept.

If we were to remove the $0<|x-a|<\delta$ condition and rewrite the definition as you proposed:

$$\lim_{x\to a}f(x)=L\iff\forall\varepsilon>0;\big(|f(x)-L|<\varepsilon\big)$$

This would mean that for every $\varepsilon>0$, the values of $f(x)$ are within $\varepsilon$ of $L$ for all $x$ in the domain of $f$. This is a much stronger condition than the original definition of a limit and is not what we want. The original definition only requires $f(x)$ to be within $\varepsilon$ of $L$ for $x$ sufficiently close to $a$ (i.e., within a $\delta$-neighborhood of $a$).

So, in conclusion, the $0<|x-a|<\delta$ condition in the $\varepsilon$-$\delta$ definition of a limit is not just trivially true, it's a vital part of the definition that captures the idea of "closeness" inherent in the concept of a limit.