Suppose we want to prove that

$$ \lim_{x \to a} f(x) = L,$$

that is

$$ \forall \epsilon > 0, \exists \delta > 0, \forall x \in \mathbb{R}, 0 < |x - a | < \delta \Longrightarrow |f(x) - L| < \epsilon.$$

Why are we constraining $|x - a|$ to be less than $\delta$ ? What happens if $0 < |x - a | \leq \delta$, or more specifically, $0 < |x - a | = \delta$ ?

Could we prove then that $|f(x) - L| \leq \epsilon$, or is there some other problem?

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    $\begingroup$ Instead you can assume $0<|x-a|\leq \delta$ $\endgroup$
    – user673903
    May 27, 2019 at 0:06
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    $\begingroup$ On $\mathbb{R}$ it doesn't matter since you can insert a "fudge factor" so $\lvert x-a\rvert\leq\delta<2\delta$ and so get away with it. Later, you will see that $<$ give you the correct idea for open sets, and $\leq$ doesn't. $\endgroup$ May 27, 2019 at 0:11
  • $\begingroup$ @Milad That's what I meant, why can't we add an "or equal" sign to that compound inequality. I mention it later in the post. Should I edit the title to make it more clear? $\endgroup$
    – Chrisuu
    May 27, 2019 at 0:30
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    $\begingroup$ Yes I just saw you’d mentioned it too. The reason is maybe clear. In topology which is a generalization of $\mathbb R$, the general notion of continuity is defined using open sets. That’s why usually the equality is not used in this context $\endgroup$
    – user673903
    May 27, 2019 at 0:33

3 Answers 3


There are two different issues here:

  1. Would it still be correct if instead of requiring a strict inequality, we allowed a non-strict inequality, with the definitions of limits, continuity, etc?

  2. If it is still correct, why do we use the strict inequality definition instead of the non-strict inequality?

Answer the first. For the real numbers, yes, it would still be “correct” to define limits (and thus, continuity) in terms of a non-strict inequality, in the sense that if we replace the strict inequality with a non-strict inequality in the definitions, then everything that satisfies the new definition also satisfies the old one, and vice versa: you get equivalent concepts.

That is, let us say that we define limits as follow:

Let $f$ be a function defined on an interval that contains $a$ in its interior, except perhaps at $a$, and let $L$ be a real number. We say that $\lim\limits_{x\to a}f(x) = L$ if and only if for every $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that for all $x$, if $0\lt |x-a|\leq \delta$, then $|f(x)-L|\leq \epsilon$.

I claim that any time you can say the limit of $f(x)$ as $x\to a$ is $L$ using this definition, you can also say it using the old (strict inequality) definition, and vice versa.

To see this, suppose that $f$, $a$, and $L$ satisfy this definition. To show it satisfies the usual one, let $\varepsilon\gt 0$. We want to show that there exists a $\delta\gt 0$ such that if $0\lt |x-a|\lt \delta$, then $|f(x)-L|\lt \varepsilon$ (the “old” definition). We know that given any $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that if $0\lt |x-a|\leq \delta$, then $|f(x)-L|\leq \epsilon$.

Well, let $\epsilon = \frac{\varepsilon}{2}$. We know there exists $\delta_1\gt 0$ such that if $0\lt |x-a|\leq\delta_1$, then $|f(x)-L|\leq \epsilon$. Let $\delta=\frac{\delta_1}{2}$. I claim that if $0\lt|x-a|\lt \delta$, then $|f(x)-L|\lt \varepsilon$ (what we want to prove). Indeed, suppose that $x$ is such that $0\lt |x-a|\lt \delta=\frac{\delta_1}{2}$; then we also have $0\lt|x-a|\leq \delta_1$, and so we know that this implies that $|f(x)-L|\leq \epsilon = \frac{\varepsilon}{2}\lt \varepsilon$. Thus, we have found a $\delta\gt 0$ such that if $0\lt |x-a|\lt \delta$, then $|f(x)-L|\lt \varepsilon$, as desired.

Conversely, if $f$, $a$, and $L$ satisfy the definition with strict inequality, then they also satisfy the new one (with the same $\delta$). Thus the two definitions are equivalent.

Answer the second. If they are equivalent (essentially, “the same”), why do we use the strict inequality definition instead of the non-strict one?

Well, that has to do with generalizations: basically, the notion of limit and continuity make sense is much more general contexts than just the real numbers. This is one of the concerns of the branch of mathematics known as Topology.

Turns out that the key concept in topology is that of open sets (well, there’s lots of ways of defining a topology, all equivalent, and one of them is the notion of open sets). For real numbers, the notion of open sets corresponds to “union of open intervals”. What we want for a function to be continuous is that if you take an open set in the range, then the collection of all points in the domain that map into the given open set is itself an open set. For real numbers, this turns out to be translated into the strict inequality version of the definition.

Thus, it makes more sense to use the strict inequality definition, because it generalizes better. And so that’s why we use it:

We use the strict inequality version because it generalizes better, and it corresponds to the notion of limit and continuity that are correct in Topology.

  • 2
    $\begingroup$ You’ll make a good instructor $\endgroup$
    – user673903
    May 27, 2019 at 1:06
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    $\begingroup$ @Milad: Given that I’ve been a professor for over 20 years, thank you, but the praise is a bit late in coming... ;-) $\endgroup$ May 27, 2019 at 1:07

We want $0<|x-a|<\delta$ because this determines a punctured neighborhood around $a$. You can also use $0<|x-a|\leq \delta$, since you have $\{x:0<|x-a|\leq \delta'\}\subset \{x:0<|x-a|< \delta\}$ for $\delta'=\delta/2$. Now, you can't use $0<|x-a|=\delta$ in your definition since this only gives information on few points approaching $a$ and not an entire neighborhood, and in fact if you use this, the limit wouldn't be unique.

For instance, suppose you have a function such that $f(1/{2n})=1$ and $f(1/(2n+1))=2$ for all $n\in\mathbb{Z}$. Then, you certainly have that $$\forall \epsilon,\exists \delta=1/2,\forall x,|x-0|=1/2\rightarrow |f(x)-1|<\epsilon$$ And also $$\forall \epsilon,\exists \delta=1/3,\forall x,|x-0|=1/3\rightarrow |f(x)-2|<\epsilon$$

As you can see, $1$ and $2$ would be "limits" with this definition.


While $0<|x-a|\le \delta$ works fine , it is more convenience to use, $0<|x-a|<\delta$ but $0<|x-a|=\delta$ does not work at all.

You want to keep $x$ in small neighborhoods of $a$ in order to have $f(x) $ in small neighborhood of $L$

With $0<|x-a|=\delta$ , your $x$ is not close enough to $a$

For example let $f(x)=|x |$ and suppose someone claims that $$\lim_{x\to 0} f(x)=1 $$

We know that the claim is false but given any $\epsilon >0$ and $\delta =1$ we have

$$|x-0|=1\implies |f(x)-1|=0<\epsilon$$

As you see we need to keep $x$ around $a$ and it is why we need $|x-a|<\delta$


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