TITLE: I feel like epsilon-delta is reversed
QUESTION [11 upvotes]: 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".

REPLY [4 votes]: "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.

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.