I feel like epsilon-delta is reversed

Question

The lim is about "when $x$ approaches $a$, then $y$ approaches $L$". 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 $-\epsilon<f(x)-L<\epsilon$? But, the conventional explanation says like "for all epsilon, you find delta", which feels like to me, "when $y$ approaches $L$, $x$ goes to $a$".

Answer 1

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.)

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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 $\varepsilon$.

Answer 2

"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.

Answer 3

Here is my two cents: I think you're confusing it with folks who usually start with the $\epsilon$ inequality and goes backwards to find the $\delta$ which depends on $\epsilon$. This is a very effective method for finding your $\delta$ but the formal proof in the end is frowned upon in the math community to be written backwards. This is the key, your choice of existing $\delta>0$ must depend on an arbitrary given $\epsilon>0$. A lot of texts will say:

"The limit of $f(x)=L$ as $x \to a$ means

$$\vert f(x) -L \vert <\epsilon.$$

whenever $\vert x - a\vert<\delta$."

I always like to put it as follows:

For any given $\epsilon>0$, there exists a $\delta>0$ such that if

$$\vert x-a\vert<\delta,$$

then

$$\vert f(x)-L\vert<\epsilon.$$

I.e., smallness is preserved. So a small piece in your domain must get mapped to a small piece in your domain.

For instance, to show for

$$f(x)=2x-4$$

that $\lim_{x \to 2}2x-4=0$, you want for any given arbitrary $\epsilon>0$ to find $\delta>0$ such that if

$$\vert x-2 \vert<\delta,$$

then

$$\vert 2x-4 \vert <\epsilon.$$

So here we can deduce that

$$\vert x-2\vert<\frac{\epsilon}{2}.$$

This means the choice of $\delta$ is $\delta=\frac{\epsilon}{2}$. So we have if

$$\vert x - 2 \vert<\delta=\frac{\epsilon}{2}.$$

I.e., after multiplying both sides by 2 and by scalar property of absolute value,

$$\vert 2x-4 \vert<\epsilon.$$

This becomes more apparent once you realize if $(X,\tau),(Y,\sigma)$ are two topological spaces, then

$$f:X \to Y$$

is said to be continuous if for every $V \in \sigma$, we have that $f^{-1}(V) \in\tau$. And for metric spaces, these definitions coincicide.