# Interpretation of $\epsilon$-$\delta$ limit definition

The epsilon-delta definition for limits states that (from Wikipedia) for all real $\epsilon > 0$ there exists a real $\delta > 0$ such that for all $x$ with $0 < |x − c | < \delta$, we have $|f(x) − L| < \epsilon$ - however, the definition of the limit requires only the existence of some $\delta>0$ for any $\epsilon>0$. The part I am having trouble understanding is why there are no details as to the "intuitive" decrease of the δ as ε grows smaller.

I realize that saying that as ε approaches zero δ also approaches zero would use the non-rigorous intuition of a limit in a definition meant to make the limit a rigorous part of mathematics, but why is it unnecessary to show the relationship between epsilon and delta besides the proof of existence? Is there some other implication of a function that I am missing that is the reason only the proof of existence is in this definition?

EDIT: If we consider the dependence of on an epsilon on a decreasing delta, can the limit exist if epsilon is increasing as delta decreases? If so, why?

• Thanks for the clarifications; I have included them in edits. As you have mentioned, δ(ϵ) tends to get smaller, but why? Are there cases where it does not become smaller, and if so, why is this definition still considered rigorous? – hedgepig Jul 24 '13 at 10:52
• Given $\epsilon>0$ there is no "well-defined corresponding $\delta>0$", not even a "maximal admissible $\delta$" which then would depend on $\epsilon$ in a unique way. When we write $\delta(\epsilon)$ we do so just to indicate that making $\epsilon$ smaller might force us to make $\delta$ smaller as well. – Christian Blatter Jul 24 '13 at 11:17
• The statement, "if $\epsilon$ decreases then the $\delta$ we can choose also has to decrease" is false in general. Consider the constant function $f(x) = 2$. Suppose we want to prove $\lim_{x\to a} f(x)=2$ for some $a$. Here no matter what $\epsilon$ we are given, we can choose our $\delta$ to be anything. Thus if $\epsilon = 0.1$, we can choose $\delta = 100$. If $\epsilon=0.01$, we can choose $\delta = 1000$. – Alraxite Jul 24 '13 at 11:32
• @Ittay Weiss: The $\sup$ of the admissible $\delta$'s could be $\infty$. But in the first place we don't want to bother about computing such a $\delta_{\rm max}$ when we don't really need it for numerical work or similar. – Christian Blatter Jul 24 '13 at 19:39
• @ChristianBlatter $\infty$ is ok in the context of the definition of limits. This notion of limit is essentially metric, and in the context of metric spaces it is more natural to consider $d:X\times X\to [0,\infty ]$. Allowing $\infty$ makes all sort of things work more smoothly, e.g., having a unique well-defined $\delta$. And in some cases it is important to analyze how $\delta$ varies when $\epsilon$ varies. – Ittay Weiss Jul 24 '13 at 19:46

For the purposes of limits, the precise dependence of $\delta$ on $\epsilon$ is simply not important. The proofs do not require any knowledge of that relation. As said, $\delta (\epsilon)$ usually tends to $0$ as $\epsilon$ tends to $0$, but this is not always the case. To make this precise, let $\delta(\epsilon)$ be the largest $\delta$ corresponding to $\epsilon$ in the definition of continuity at $x$ for a function $f$. Thus, $\delta(-)$ is a function whose domain is $(0,\infty )$ and whose range is $(0,\infty]$, and it is monotonically non-decreasing. If $f$ is a constant function, then $f(\epsilon)=\infty$ for all $\epsilon>0$, showing that indeed $\lim_{\epsilon\to 0}\delta(\epsilon)$ need not be $0$.
The definition of limit captures the following: $\lim_{x\to a}f(x)=L$ means that for any prescribed distance $\epsilon>0$, there exists some upper bound for distances $\delta$ such that if $x\ne a$ is within $\delta$ units from $a$, then $f(x)$ is guaranteed to be within $\epsilon$ units from the limit $L$.
Remark: The function $\delta(-)$ above is known as a modulus of continuity for $f$. Functions whose moduli of continuity have certain properties (e.g., are concave) are of importance.
• Let $\delta (\epsilon)$ be the largest positive number such that if $0<|y-x|<\delta (\epsilon)$, then $|f(y)-L|<\epsilon$. – Ittay Weiss Jul 24 '13 at 11:41
• @CameronBuie in meant 'positive extended real number'. In my answer I mention the constant functions, and I mention that the codomain of $\delta(-)$ is $(0,\infty ]$. – Ittay Weiss Jul 24 '13 at 19:28