The lim is about "when x approachs a, then y approachs L". Then, shouldn't the epsilon and delta be like "For all delta, no matter how small the delta is, you can always find an epsilon that makes ε < f(x)-L < ε"? But, the conventional explanation says like "for all epsilon, you find delta", which feels like to me, "when y approachs L, x goes to a".

  • $\begingroup$ You meant "-ε < f(x)-L < ε", right? "ε < ε" is always false. $\endgroup$ Apr 21, 2022 at 8:10
  • $\begingroup$ You've got a good answer, but if it helps, the proposed condition is equivalent to "$f$ is bounded." As the initial quantifier, "for all $\delta > 0$" is harder to meet for large $\delta$ so it incentivizes "picking $\delta>0$ as large as possible." Similarly, putting "there exists an $\varepsilon>0$" second incentivizes "picking $\varepsilon$ as large as possible." $\endgroup$ Apr 21, 2022 at 12:21

2 Answers 2


This is a common misunderstanding, and the only response I can ever think of is to just examine what the definition is really getting at.

The statement $\lim\limits_{x\to a}f(x)=L$ means that, when $x$ is close to $a$, $f(x)$ is close to $L$. So we want something like "if $|x-a|$ is small, then $|f(x)-L|$ is small." But then we have to decide what "small" means. If $|x-a|$ is smaller than some $\delta>0$, how small should $|f(x)-L|$ be?

The answer is that we want to ensure $|f(x)-L|$ is arbitrarily small as long as $x$ is sufficiently close to $a$. This means that, if we want $f(x)$ within $0.001$ of $L$, I can tell you how close $x$ must be to $a$. And there is no reason to find the optimal choice for this distance; the important thing is that there is some $\delta>0$ for which $|x-a|<\delta$ is enough to guarantee $|f(x)-L|<0.001$. This should work not just for $\varepsilon=0.001$, of course, but for any $\varepsilon>0$.

(By the way I should be saying $0<|x-a|<\delta$ but you get the point.)

It's also worth examining why the reverse definition is not what we want. We could try saying "for any $\delta>0$ there exists $\varepsilon>0$ such that $0<|x-a|<\delta$ implies $|f(x)-L|<\varepsilon$." The immediate issue is that $\varepsilon$ might always be huge. In the correct definition, the "for all $\varepsilon>0$" includes every arbitrarily small value for $\varepsilon$. But in the reverse definition, the smallness of $\delta$ doesn't guarantee the smallness of $\varepsilon$.

  • 2
    $\begingroup$ I've seen the reverse before in various places and I've failed to properly articulate exactly why it's not sensible. This answer is phrased very well. I think I'll steal some of the wording next time I get asked that question. $\endgroup$ Apr 21, 2022 at 2:50

"Epsilon-Delta" is supposed to be a challenge, in order to prove that $\lim\limits_{x\to a}f(x)=L$.

Your proposal

For all delta, no matter how small the delta is, you can always find an epsilon that makes -ε < f(x)-L < ε

Your proposal isn't much of a challenge. For any $\delta$, you can simply pick $\varepsilon = 2$, and you can "prove" that sine converges to $L = 0$ at every point.

sine and -2<y<2

And for any $\delta$, if you pick $\varepsilon = 10$, you can even "prove" that sine converges to $L = 5$ everywhere:

sine and -5<y<15

Limit definition

"For all epsilon, you find delta" describes the limit definition reasonably well.

"For all $\varepsilon$" actually means "However small $\varepsilon$ is", because large values aren't much of a challenge.

  • For $\left\{ \begin{array}{c} a = 0 \\ L = 0 \\ ε = 2 \end{array} \right. $, can you find a corresponding $\delta$? Sure! Any delta will do, for example $\delta = \pi$:

enter image description here

  • For $\left\{ \begin{array}{c} a = 0 \\ L = 0 \\ ε = 0.5 \end{array} \right. $, can you find a corresponding $\delta$? That's a bit harder, but $\delta = \frac{\pi}{6}$ should do:

enter image description here

The challenge gets progressively harder the smaller ε is. If a corresponding $\delta$ can always be found, and the curve is always completely included in the drawn rectangle $$\left\{ \begin{array}{c} a - \delta < x < a + \delta \\ L - \epsilon < y < L + \epsilon \end{array} \right. $$ then $\lim\limits_{x\to a}f(x)=L$.

If you pick $\left\{ \begin{array}{c} a = 0 \\ L = 0.1 \\ \end{array} \right. $, you'll soon find an ε for which no $\delta$ fits.

enter image description here

enter image description here

If you're interested, here's the corresponding interactive diagram.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .