# I feel like epsilon-delta is reversed

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".

• You meant "-ε < f(x)-L < ε", right? "ε < ε" is always false. Apr 21 at 8:10
• 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." Apr 21 at 12:21

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 $$\varepsilon$$.

• 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. Apr 21 at 2:50

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

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.

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

## 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 $$\delta = \pi$$:

• 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:

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.

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