Limit of a function with no connection between $\epsilon$ and $\delta$

Question

Which functions do not have a connection between $\epsilon$ and $\delta$ when defining the limit?

I thought about any function of the format of $f(x)=a$ when $a\in \mathbb{R}$.

Answer 1

I think this is one of the fundamental problems with modern analysis/calculus textbook which focus so much on symbolism ($\epsilon, \delta$) and so less on rigor/concepts/fundamentals.

In the definition of limit what is required is to find a $\delta > 0$ corresponding to any $\epsilon > 0$ such that some inequalities containing $\epsilon$ hold whenever some inequalities containing $\delta$ hold.

It is not necessary to find $\delta$ as a unique function of $\epsilon$ (i.e. write $\delta$ as some expression containing $\epsilon$) for the simple reason that if one value of $\delta$ works any smaller value also works. The more important aspect is that it should be possible to find $\delta$ for any $\epsilon$ so that the argument for finding $\delta$ should not be dependent on any particular value/property of $\epsilon$ apart from the only requirement that $\epsilon$ be positive. Thus there should not be any value of $\epsilon > 0$ for which it is impossible to find a desired $\delta$. This is the actual meaning of phrase "find a $\delta > 0$ corresponding to any arbitrary $\epsilon > 0$".

The connection between $\epsilon, \delta$ is not supposed to be a direct one like a function. The definition of limit rather gives us a much more liberty in finding $\delta$ based on $\epsilon$ rather than restricting with functions or exact formulas which could express $\delta$ in terms of $\epsilon$.

I guess the one line answer to your question is as follows:

Limit definition does not require a connection between $\delta$ and $\epsilon$ but rather needs the existence of at least one value of $\delta > 0$ corresponding to any $\epsilon > 0$.

Answer 2

If $f$ is defined to be $a$ for all $x \in \mathbb{R}$, the limit as $x \to b \in \mathbb{R}$ of $f(x)$ is going to be $a$, i.e. $|x-b|<\delta \implies |f(x)-f(b)|=|a-a|=0<\epsilon$ is always true for all $\epsilon>0$.

Answer 3

I think there may be a conceptual misunderstanding underpinning your question as it currently stands. To see this, consider the following common definition of a limit:

Let $f$ be a function defined on some open interval that contains the number $a$, except possibly at $a$ itself. Then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$, and we write $\lim_{x\to a}f(x)=L$ if for every number $\epsilon>0$ there is a number $\delta>0$ such that $|f(x)-L|<\epsilon$ whenever $0<|x-a|<\delta$.

To your point specifically:

I thought about any function of the format of $f(x)=a$ when $a\in \mathbb{R}$.

There's definitely a connection between $\epsilon$ and $\delta$ here per the definition of a limit, but perhaps that connection is not all that apparent. Perhaps a simple example will help.

Example 1: Consider $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=c$. Prove that $\lim_{x\to a}f(x)=c$ using the $\epsilon$-$\delta$ definition of a limit.

Proof. Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-a|<\delta$, then $|c-c|<\epsilon$. But, $|c-c|=0$; thus, this will be true no matter what $\delta$ we pick. $\blacksquare$

Sure, the connection between $\epsilon$ and $\delta$ may not be all that palpable here since $\delta$ can be whatever you want, but that does not change anything concerning the relationship between $\epsilon$ and $\delta$ in the definition of the limit. Another example where the connection between $\epsilon$ and $\delta$ is simple yet possibly somewhat subtle is as follows.

Example 2: Consider $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=x$. Prove that $\lim_{x\to a}f(x)=a$ using the $\epsilon$-$\delta$ definition of a limit.

Proof. Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-a|<\delta$, then $|x-a|<\epsilon$. Thus, $\delta=\epsilon$ will work. $\blacksquare$

Does that clear things up at all?