# Struggling to understand epsilon-delta

The definition of a limit is:

$\lim_{x\to a}f(x)=L$ if for every $\epsilon > 0$ there is a $\delta > 0$ so that whenever $0 < \lvert x - a \rvert < \delta$ we have $\lvert f(x) - L \rvert < \epsilon$

Now it seems pretty intuitive. But I am hung up on a few problems:

1. Many pictures show something like this:

epsilon-delta

This seems intuitive at first and it demonstrates that $\lvert x - a \rvert$ and $\lvert f(x) - L \rvert < \epsilon$ are not necessarily the same (as the graph can be deceptive, especially if $f(x)$ is a straight line) as when you are projecting from $L$ to the graph down to $a$, $\lvert x - a \rvert$ and $\lvert f(x) - L \rvert$ will be different. The problem in my understanding became apparent when I saw a similar graph in a textbook where the projected lines were not $\lvert f(x) - L \rvert$, but projection for aesthetic purposes and that it was bounded by $\lvert f(x) - L \rvert$. I then realized I don't get it geometrically at all (Google "mooculus", page 20).

2. I don't understand what the "verification" in the proof is. It seems to be a tautology. For example take $\lvert f(x) - L \rvert < \epsilon \Longrightarrow \lvert (3x - 1) - 2 \rvert < \epsilon$. You will eventually get to $\lvert x - 1 \rvert < \epsilon/3$. Then the proof is "completed" by showing that $\lvert x - a \rvert < \delta \Longrightarrow \lvert x - 1 \rvert < \epsilon/3 \Longrightarrow \lvert f(x) - L \rvert < \epsilon$. But $\delta$ is taken to be $\epsilon/3$. It seems to be the equivalent of demonstrating that $x + 1 = 2$ by plugging in $-3$ into it.

3. The proof starts by either assuming the limit exists or doesn't exist. In fact, I've found many textbooks or teachers to take this approach:

Think of it as a game. You give me an $\epsilon > 0$ and I can give you a $\delta > 0$...

But they typically omit a glaring part: that if this doesn't hold, the limit doesn't exist. Also glaringly missing, I haven't seen an example of a proof that isn't simply "proving" or "showing" the premise that is already assumed! How would you for example use the epsilon-delta definition to show a limit doesn't exist if you don't already know in advance it is the case?

To extend on number 3, I'm aware that you must choose an $\epsilon$ and that if you prove it for one, you prove it for all. However, the catch is in cases where $\lvert x - a \rvert$ needs to be restricted (i.e, $\lvert x - a \rvert < 1$) $1$ is typically chosen but this does not work in all cases! I have no intuition on how to choose an $\epsilon$ let alone know if I'm simply doing something wrong or if the limit does not exist. That is, proving the negative seems more difficult.

Can someone explain it in a different way? I've resorted to many different .edu sources, free online textbooks and even questions on this site and the pedagogy doesn't seem to be reaching me.

• There are many different answers to the same question you are posting here, on this site, try making a little search :) Dec 25, 2016 at 3:39
• Calculus books would do a great service to their readers if they gave examples how an $\epsilon - \delta$ attempt would fail if the limit didn't exist. But I haven't found many books that do. Dec 25, 2016 at 3:40
• @GFauxPas, for some reason Calculus books are often terrible and spend 10,20 or more pages in less important and surely more understandable stuff, but they never slow down where it is actually necessary. Don't know if depends on a lack of teaching ability, lack of deep knowledge or just lack of knowledge about where beginners actually struggle (which is clearly not de morgan rules, but rather delta-epsilon proofs) Dec 25, 2016 at 3:45
• @GFauxPas The OP gave an example for a textbook with an example demonstrating discontinuity in problem 1: mooculus page 20. I have to admit that they just prove discontinuity. They do not show a failing attempt to prove continuity. Do you mean that would be advantageous for educational purposes? (This is actually not too much different from proving discontinuity.) Dec 25, 2016 at 4:05
• For your 3rd point, note that definition of a limit acts as a test to check whether a specific number $L$ is or is not the limit of a function. It does not directly answer the question : does the limit exist? Hence the existence of the limit and the number $L$ have to be guessed/known by other methods. The definition is meant to be used in theoretical contexts and not as a tool to evaluate limits. Dec 25, 2016 at 5:17

This is not a complete answer but only addressing the comment of user GFauxPas which in turn addresses problem 3 of the original poster:

I've found many textbooks or teachers to take this approach:

Think of it as a game. You give me an ϵ>0 and I can give you a δ>0 ... But they typically omit a glaring part: that if this doesn't hold, the limit doesn't exist.

We show how the guessing of the right $\delta>0$ may fail at discontinuities by a slight modification of the OP's example: \begin{align*} f(x) :=\begin{cases} 3x-1&@x\neq 1\\ 0 &@ x=1 \end{cases} \end{align*} We assume that $f$ would be continuous at $x=1$. Let $\varepsilon>0$ be arbitrarily (small). Searching for an appropriate $\delta>0$ with $|f(x)-f(1)|<\epsilon$ for all $|x-1|<\delta$ we determine the pre-image of the $\varepsilon$-neighborhood $(-\varepsilon,\varepsilon)$ of $f(1)=0$: \begin{align*} \{x\in\mathbb{R}:\,|f(x)-f(1)|<\varepsilon\} &= \underbrace{\{1\}}_{\text{for }f(x)=0}\cup \{x\in\mathbb{R}\setminus\{1\}:\,|3x-1|<\epsilon\}\\ &=\{1\} \cup \left(\frac{1-\varepsilon}3,\frac{1+\varepsilon}3\right) \end{align*} Starting from $\varepsilon=2$ the open interval $\left(\frac{1-\varepsilon}3,\frac{1+\varepsilon}3\right)$ becomes disconnected to the location $x=1$ for shrinking $\varepsilon$. So the argument $x=1$ which maps into the open $\varepsilon$-neighborhood of $f(1)$ becomes insulated in the pre-image of this neighborhood. The pre-image is not a neighborhood of $x=1$ and therefore does not contain the interval $(1-\delta,1+\delta)$ for any $\delta>0$.

The following figure shows the pre-image of $(f(1)-\varepsilon,f(1)+\varepsilon)$ for $\varepsilon=1$, i.e., $f^{-1}((-1,1))=\{x\in\mathbb{R}:\, -1<f(x)<1\}$. We take $\varepsilon=1$ instead of $\varepsilon=2$ to make the insulation of $x=1$ in the pre-image of $(f(1)-\varepsilon,f(1)+\varepsilon)$ more visible. The broken circle ( ) at $(x,y)=(1,2)$ indicates that there is a gap in the linear graph at $x=1$. This gap is filled by $(x,y)=(1,0)$ as it is indicated by the bullet • there.

The blue strip indicates the y-interval $(-1,1)$ which is mapped to the pre-image $f^{-1}((-1,1))$ which is colored yellow. We take special note of the yellow ring around the bullet at $(x,y)=(1,0)$ which is the indicates point of the pre-image.

There is no room around this point for a $\delta$-interval. Furthermore, a comment on the original posters statement:

I'm aware that you must choose an ϵ and that if you prove it for one, you prove it for all.

That is not true. You must show that for all $\varepsilon>0$ you find $\delta>0$ such that $|f(x)-f(a)|<\varepsilon$ provided $|x-a|<\delta$. You do that by keeping $\varepsilon$ as a free variable and by not binding $\varepsilon$ to any specific number. Only when you disprove continuity it is sufficient to provide a specific $\varepsilon>0$ such that the pre-image of $(f(a)-\varepsilon,f(a)+\varepsilon)$ is not a neighborhood of $a$.

You're correct ... it's a game. I give an $\epsilon>0$, and then your job is to find some $\delta >0$ such that $|f(x) - L|< \epsilon$ whenever $|x-a|< \delta$. If you can always do this, then you win, and the limit exists. Saying it another way, you win if you can always make $|f(x) - L|$ as small as you like (specifically, smaller than the $\epsilon$ I gave you) by choosing a suitable $\delta$.

If I can find an $\epsilon$ for which you can't find a corresponding $\delta$, you lose -- the limit does not exist.

The unfair part is that you have to win an infinite number of rounds. If you win one round (by finding a good $\delta$), I have the right to choose a new $\epsilon$, and we have to play another round.

So, the only way for you to win, really, is to invent some procedure that automatically produces a winning $\delta$ no matter what $\epsilon$ I choose. Typically, your $\delta$ will be given by some rule that depends on the $\epsilon$ I give you. Not a formula, necessarily, but at least some well-defined process.

Let's try a couple of examples:

Example 1:

$f(x) = 2x; \; a = 1; \; L = 2$.

Easy for you to win. Given my value of $\epsilon$, you just choose $\delta = \tfrac12\epsilon$ or even $\delta = \tfrac13\epsilon$. With this procedure for choosing $\delta$, you win every round, so the limit exists. If you draw a picture, you can probably see that the factor $\tfrac12$ that appears in the winning strategy is related to the fact that the function has a slope of $2$. This kind of strategy works provided the function doesn't have an "infinite" slope.

Example 2:

$f(x) = 0$ when $x < 5$ and $f(x) = 6$ when $x \ge 5;\;$ $a = 5;\;$ $L = \text{anything}$.

The value of this function "jumps" from $0$ to $6$ at $x=5$. With this function, you can't win. If I choose $\epsilon = 2$, for example, you're dead. Why did I choose $\epsilon = 2$? Because the function value "jumps" by $6$, and I need a number smaller than $6/2$, for reasons explained below. Then, no matter how small you choose your $\delta$, the interval $[5-\delta, 5+\delta]$ will include some points $x$ where $f(x)=0$, and some points $x$ where $f(x)=6$. Your best shot at proving a limit is to try $L= 3$ (halfway between $0$ and $6$). But even this won't work. No matter how small you choose your $\delta$, there will be points $x$ with $|x-5|<\delta$, but where $|f(x) - 3|> 2$. So, in this case, there is no limit at $x=5$. You run into this losing situation whenever the function has a "jump" in its value at $x=a$. A "jump" is one form of infinite slope; there are other more exotic forms, too.

The $\epsilon$-$\delta$ approach is generally used to provide a rigorous proof that some given number is the limit. But, as you noted, you first have to guess that a limit exists, and you have to guess its value, and the $\epsilon$-$\delta$ stuff is of little help in this regard. You need intuition, guesswork, sketches of graphs, and any other tricks that work for you. Calculus is relatively easy; developing intuition is harder, and is harder to teach, too.