# Why does the delta depend on the epsilon in the definition of the limit of a function?

For example, why can't it be a definition of the limit?

$$\lim_{x \to a}f(x)=l\Leftrightarrow( \forall \delta> 0,\exists\varepsilon > 0:|f(x)-l|<\varepsilon \Rightarrow|x-a|<\delta )$$

• 1) Your definition is wrong. 2) In what you wrote, $\delta$ is independent of $\varepsilon$ however, $\varepsilon$ depend on $\delta$.
– Surb
Oct 16, 2022 at 9:20
• I voted to close because such a question begining with "Why" appeals opinion-based answers. (I understood the question as: "Why was not the proposed "non-definition" chosen, instead of the (hopefully obviously non-equivalent) traditional one?"). I hesitated to flag it as a "very low quality question", i.e. unlikely to be salvageable through editing hence to be removed. Oct 16, 2022 at 9:25
• I upvoted because the question is a profound one. The OP (i.e. original poster) is attempting to expand his intuition. Personally, I consider intuition to be critically necessary in the study of Mathematics. Oct 16, 2022 at 9:39
• Note that if you take $f(x) = x^2$, by your definition $\lim\limits_{x \to 2} f(x) = 4$ does not hold because not only does neighborhood of $2$ on the x-axis maps to neighborhood of $4$ but also neighborhood of $-2$. So $f(x)=x^2$ being close to $l=4$ does not imply $x$ being close to $a=2$
– Lab
Oct 16, 2022 at 9:42
• @AnneBauval There are ~ 250,000 why-questions. Oct 16, 2022 at 12:30

## 2 Answers

To answer your question, you have to understand how Mathematicians have come to understand and define the notion of a function approaching a limit $$L$$, as the variable used in the function approaches a specific element (e.g. the value $$a$$).

First of all, I specified that the variable [i.e. $$x$$, used in the function, i.e. $$f(x)$$] is approaching a value (i.e. $$x$$ is approaching the value $$a$$).

Note that the function $$f(x)$$ might not even be defined at the value $$a$$. I think that the easiest way to understand what is happening is to think in terms of neighborhoods.

The idea is that if you specify a value $$\epsilon$$, you are requiring that for all $$x$$ in the domain of the function such that $$x$$ is in a certain neighborhood, that the distance of the value $$f(x)$$ from the limit $$L$$ will be less than $$\epsilon.$$

The variable $$\epsilon$$ is nothing more than an arbitrarily named variable. The Mathematical requirement that $$f(x)$$ is within $$\epsilon$$ of the limit $$L$$ is expressed as

$$|f(x) - L| < \epsilon. \tag1$$

The notion of $$x$$ approaching $$a$$ can be Mathematically expressed as

$$0 < |a - x| < \delta. \tag2$$

Actually, the expression in (2) is something of an oversimplification. Actually, what is required is that (1) above holds, whenever (2) above holds and $$x$$ happens to be in the Domain of the function $$f(x)$$.

I mention this complication only to warn you that the complication exists. At this point, I advise you to ignore such a complication. Instead, simply focus on the idea that the notion of $$f(x)$$ approaching a limit $$L$$, as $$x$$ approaches the value $$(a)$$ informally represents that:

For any $$\epsilon$$ that is chosen, a value $$\delta$$ can be found so that if the expression in (2) above is true, then the expression in (1) above is true.

Note that the value $$\delta$$ in such a circumstance may well depend on the value $$\epsilon.$$ In fact, you may see a reference to $$\delta(\epsilon)$$ or $$\delta_\epsilon$$ to emphasize that as the value of $$\epsilon$$ changes, the value of $$\delta$$ may change.

Put simply, the idea is that under the assumption that the limit $$L$$ exists for $$f(x)$$, as $$x$$ appproaches the value $$a$$, for any arbitrary value of $$\epsilon$$ a $$\delta$$-neighborhood can be found so that (1) is satisfied whenever (2) is satisfied.

This somewhat obtuse Mathematical notion of the limit of a function was developed to facilitate studying the behavior of functions.

It is important to examine the first portion of (2) above, closely. The requirement $$0 < |x - a|$$ indicates that as far as studying the limit of a function, there is no requirement that the function exists at the value $$(x = a)$$. Further, even if $$f(a)$$ does exist, there is no requirment that $$L$$ equals $$f(a)$$.

As an illustrative example, consider the function $$f(x)$$ defined by:

• $$f(0) = 1$$

• $$f(x) = x ~: ~x \neq 0.$$

Then, the limit $$L$$, as $$x$$ approaches $$(0)$$ of $$f(x)$$ is $$0$$. This is despite the fact that $$f(0)$$ equals $$(1)$$. In fact, this specific function is an example of a function that has a limit, as $$x$$ approaches $$(0)$$, but where the function is not continuous at $$x = 0$$.

Informally, you can regard continuity (or the lack of continuity) as involving whether you can draw the function on a piece of scratch paper, at a specific value (i.e. $$x = 0$$) without picking up your pencil.

Although this definition isn't very precise, it does provide an informal intuition about the whole notion of continuity, and it does illustrate the difference between the limit of a function existing as the variable $$x$$ approaches a certain value (i.e. $$x=0$$) and the function being continuous at ($$x = 0$$).

The limit does exist, for the bullet-pointed function $$f(x)$$ above, as $$x$$ approaches $$0$$. As a demonstration of this, consider what happens when you set

$$\delta(\epsilon) = \epsilon.$$

Then, for that specific function, as $$x$$ approaches $$0$$, whenever the inequality in (2) above is satisfied, the inequality in (1) above will be satisfied.

As a further example, consider the function :

• $$g(x) = 2x.$$

As $$x$$ approaches $$(0)$$, $$g(x)$$ approaches $$0$$. The $$\epsilon, \delta$$ relationship represented in (1) and (2) above, may be established in this situation, for the function $$g(x)$$, by specifying (for example)

$$\delta = \frac{\epsilon}{2}.$$

One answer to OPs proposal of a function limit could be it does not met the expectations we usually have of it. Let's consider for instance the constant function \begin{align*} &\color{blue}{f:\mathbb{R}\to\mathbb{R}}\\ &\color{blue}{f(x)=l} \end{align*} and let's assume we want to calculate $$\lim_{x\to 0}f(x)$$, the limit of $$f$$ at $$a=0$$.

Here we might want to say: Yes, of course, the constant function $$f(x)=l$$ has an accumulation point at $$a=0$$, the function limit at $$a$$ exists and $$\lim_{x\to 0}f(x)=l$$ of course. In order to justify this statement we recall a common definition of a function limit.

A common definition:

• Let $$f$$ be a function with domain $$\mathbb{R}$$ and $$a$$ be an accumulation point of its domain. We define \begin{align*} \lim_{x\to a}f(x)=l\quad\Leftrightarrow\quad &\forall x\in \mathbb{R}\,\forall \varepsilon >0\, \exists\delta>0: \\ &0<|x-a|<\delta \Rightarrow |f(x)-l|<\varepsilon\tag{1} \end{align*}

• We observe when comparing it with OPs proposal, that besides reordering of quantors and conditions we also want to provide some information about proper candidates for limit points and about the players $$x$$ which are allowed to be used approaching a limit.

• Possible limit points like $$a$$ above need not necessarily be part of the domain of $$f$$ but at least they have to be an accumulation point of the domain.

• Possible candidates $$x$$ approaching to the limit point are all points of the domain, which are preferrably near to the limit point, since they have to met the condition $$0<|x-a|<\delta$$.

It's not hard to show $$\lim_{x\to 0}f(x)=l$$. Given $$\varepsilon >0$$ we can take an interval around $$a=0$$ of any size we like, since due to the constant nature of $$f$$ the image is always $$f(x)=l\in (f(0)-\varepsilon,f(0)+\varepsilon)=(l-\varepsilon,l+\varepsilon)$$. So we can simply take $$\delta=\varepsilon$$ and we are finished.

OPs proposal:

Now we consider OPs proposal \begin{align*} \lim_{x \to a}f(x)=l\Leftrightarrow( \forall \delta> 0,\exists\varepsilon > 0:|f(x)-l|<\varepsilon \Rightarrow|x-a|<\delta ) \tag{2} \end{align*} and check, if according to this definition the constant function $$f(x)=l$$ has a limit at $$a=0$$.

• Since we have to find for all $$\delta > 0$$ an $$\varepsilon>0$$ we set $$\delta =1$$ and check if we can find an admissible $$\varepsilon>0$$.

• We want to find an $$\varepsilon>0$$ which mets the condition $$\color{blue}{0<|f(x)-l|<\varepsilon \Rightarrow|x-a|<\delta}$$ in (2). But, whichever $$\varepsilon>0$$ we choose, again due to the nature of the constant function $$f(x)=l$$ we can always find $$x$$ far outside the interval $$|x-a|=|x|<\delta$$ but with \begin{align*} |f(x)-l|<\varepsilon \end{align*} so that the condition does not hold for all $$x\in\mathbb{R}$$ for which $$|f(x)-l|<\varepsilon$$ is true.

Conclusion: Since OPs proposal does not met some common cases we want to cover, it is not that useful.

Note: Interestingly the definition of a function limit is not unique. There are two different definitions commonly used. See this post for some information.