Epsilon-Delta Proof Of a Function - Do Epsilon and Delta decrease (resp. increase) With Respect to Each Other?

Question

I am a first year Mathematics student. And during a lecture we were taking a look at the limit of a function and I asked my lecturer if in the ε- proof of limit of a function, can we always say that if ε is increasing(respectively decreasing), so does . In other words, if ε is increasing(respectively decreasing), does always increase(respectively decrease)? And we couldn't really be sure about it in the lecture so I wanted to ask it here, to everyone.

I am not sure about which tags I should've put for this question so let me know if this question is related to some tag(s) which I didn't include for this question.

Any kind of help is very appreciated.

Answer 1

The $\varepsilon-\delta$ limit only requires the existence of a $\delta$, it doesn't put any restrictions on it. Perhaps most importantly, it doesn't require that $\delta$ is unique.

As a bit of an extreme example, consider the constant function $f(x) = c$ and let's look at what happens in the neighbourhood of $x_0 = 0$. Of course we already know that $\lim_{x \rightarrow 0} f(x) = c$, but notice that for the $\varepsilon-\delta$ proof that for any $\varepsilon > 0$ we can choose any positive $\delta$ we like, and in particular we could make $\delta$ increase as $\varepsilon$ decreases and our proof will still work.

For example, let $\delta = \frac{1}{\varepsilon}$. Then for $|x - 0| = |x| < \delta$, $|f(x) - f(0)| = |c - c| = 0 < \frac{1}{\delta} = \varepsilon$.

Of course we could also choose a tighter bound on $\delta$, and in general you will find that usually the $\delta$ will be mostly non-increasing with decreasing $\varepsilon$, but it's not a fundamental property.

Answer 2

Well with not much thinking, I have found another counterexample for it. And for beginners like me, I tried to visualize it. This purple curve is a graph of a function and see, as ε decreased, increased instead. I hope I haven't made any mistakes on this.

Also, I made it on Microsoft Paint so I am open for any suggestions for mathematical drawings.

Answer 3

The first answer you received is correct and demonstrates why $\delta$ does not need to increase or decrease as $\varepsilon$ does.

I think part of the reason for the confusion on this point in the first place is the use of an inappropriate graph of the relationship of $\delta$ and $\varepsilon$ to each other and to the function whose limit you want to take.

I don't know where that graph came from (not from a textbook, I hope!) but I would propose to modify the graph as follows:

The difference is that the portion of the function bracketed by the $\delta$ intervals is strictly inside the portion of the function bracketed by the $\varepsilon$ intervals, so that neither the upper line $y=L+\varepsilon$ nor the lower line $y=L-\varepsilon$ touches the portion of the function graph between $x=c-\delta$ and $x=c+\delta.$

If you study this diagram carefully it should be relatively easy to understand that you can increase this value of $\varepsilon$ as much as you want without having to change $\delta$ at all. Conversely you can make $\delta$ as small as you want without changing $\varepsilon.$

The original diagram simply does not work in general, because for a given $c$ and $\varepsilon$ it is not generally possible to find $\delta$ such that $f(c-\delta) = L-\varepsilon,$ $f(c) = L,$ and $f(c+\delta) = L+\varepsilon.$ For an extreme example of this, consider the case where $f(c)$ is the maximum value of the function. The line $y=L+\varepsilon$ will pass clear over the maximum point of the function without touching the curve anywhere, which makes it obvious that you can't have $f(c+\delta) = L+\varepsilon.$

What is far more important than how close $f(c+\delta)$ is to $L+\varepsilon$ is the requirement that all of the graph of the function between $x=c-\delta$ and $x=c+\delta$ must lie between the lines $y=L+\varepsilon$ and $y=L-\varepsilon.$ If the function ever wanders outside the region between those two horizontal lines while traveling between the two vertical lines, that value of $\delta$ is invalid.