A terminology to analysts When analysts say "$\epsilon$ (or whatever greek symbol) can be chosen arbitrary small", do they really just mean we can take $\epsilon = 0$ or $\epsilon \to 0$ later?
When I asked myself this question, I immediately began doubting my understanding of limits.
That is $\forall \epsilon>0 ,\exists \delta > 0 :|x-a|<\delta \implies |f(x) - L|<\epsilon$.
I was taught that a very long time ago, a limit is something you approach very closely, but not exactly equal to. So what logic rule am I breaking if i take $\epsilon = |f(x) - L|$?
Also, what is the advantage of proving two things are equal by saying they are epsilon close to each other? Isn't this really just make it harder than it needs it to be?
EDIT
I have also noticed that some other definitions that uses epsilon distance instead of just saying they are equal to each other.

$f \in R(\alpha)$ on $[a,b] \iff \forall \epsilon >0, \exists$ partition $P$ such that $$U(f,P,\alpha) - L(f,P,\alpha) < \epsilon$$

Now what is wrong with saying

$f \in R(\alpha)$ on $[a,b] \iff  \exists$ partition $P$ such that $$U(f,P,\alpha) =  L(f,P,\alpha) $$

 A: More correctly, if $f(x)$ is a real-valued function defined in a real open interval around a real number $a$ (except possibly at $a$) then we say $\lim_{x\to a}f(x)=L$ if $$\forall\epsilon>0,\exists\delta>0:0<|x-a|<\delta\implies|f(x)-L|<\epsilon.$$
That $\epsilon$-$\delta$ language means that for any positive $\epsilon$, we can keep $f(x)$ within a distance $\epsilon$ of $L$ simply by making sure that $x$ stays within some distance $\delta>0$ of $a$ (except perhaps if $x=a$). Eliminating the $\epsilon$ and $\delta$ entirely, this means that we can keep $f(x)$ "as close as we like" to $L$, simply by keeping $x$ "close enough" to (but not equal to) $a$.
How close is "close enough"? Well, that will depend on several factors in general, and almost always on how close we'd like $f(x)$ to stay to $L$--if we want to keep $f(x)$ closer to $L$, we may need to keep $x$ closer to $a,$ too.
Note that $x$ is still a variable in this definition--unlike $L$ (and $a$), it is not fixed, so we can't simply take $\epsilon=|f(x)-L|.$
Now, if we fixed some $x_0\ne a$ and happened to know that $f(x_0)\ne L,$ then we could certainly take $\epsilon=|f(x_0)-L|.$ Then if $\lim_{x\to a}f(x)=L,$ we'd know that there was some $\delta>0$ such that $|f(x)-L|<\epsilon$ whenever $0<|x-a|<\delta$--that is, we could keep $f(x)$ closer to $L$ than $f(x_0)$ is, by making sure that $x$ stays within $\delta$ of (but not equal to) $a$. Note that such a $\delta$ will necessarily be less than $|x_0-a|,$ since if not, then $x_0$ is an $x$ that is within $\delta$ of (but not equal to) $a$, yet $|f(x_0)-L|=\epsilon$.
Let me sum up the discussion the previous paragraph: If there is some $x_0\ne a$ with $f(x_0)\ne L$, and we know that $\lim_{x\to a}f(x)=L,$ then we can keep $f(x)$ closer to $L$ than $f(x_0)$ is, by making sure that we keep $x$ closer to (but not equal to) $a$ than $x_0$ is by a sufficient amount.
Hopefully that summary doesn't surprise you or confuse you, but let me know if it does.
Does this help clear things up for you?
A: They certainly do not mean you can take $\epsilon=0$. Usually, the statement

$\epsilon$ can be chosen arbitrarily small

means that whatever property the statement is referring true holds for any $\epsilon>0$. Often this fact is used to take some limit as $\epsilon\to 0^+$ (called the limit from the right as $\epsilon$ goes to $0$).
A: This is really more of an extended comment, rather than an answer.
From the point of view of pure logic, the phrase "$\epsilon > 0$ can be chosen arbitrarily small" simply means "for all $\epsilon > 0$."  That's it.
So why include the "arbitrarily small" part?
The phrase "arbitrarily small" is simply telling the reader how s/he should think about the property involving $\epsilon$.  Namely, it tells the reader that it will be relatively easy to satisfy the property with large values of $\epsilon$, but harder to satisfy it with smaller values of $\epsilon$.  And yet, despite this difficulty when small values of $\epsilon$ are taken, the statement is still true!

Example: Suppose I tell you that for every $\epsilon > 0$, there exists a real number $\delta$ satisfying $0 < \delta^3 < \epsilon$.  You're not sure whether to believe me, so you try examples:

*

*If $\epsilon = 1$, then there are lots of numbers $\delta$ that make the statement true.  (e.g. every $0 < \delta < 1$ works)


*If $\epsilon = 100$, then there are even more values of $\delta$ that make the statement true. (e.g. every $0 < \delta < 4$ works)
In fact, if you've proven that my statement is true for a given $\epsilon$, then it's also true for all greater $\epsilon' > \epsilon$.  Like, if it's true for $\epsilon = 1$, then it's true for $\epsilon = 2.5, 17, 10000$, etc.
But I said that it's true for all $\epsilon > 0$.  That means that we have to look at small values of $\epsilon$, too.  And in these cases, values of $\delta$ that work are less plentiful.
In such a situation, I would say that my statement is true for all $\epsilon >0$, no matter how small, to emphasize this point.
Note, however, that although my statement is true for all $\epsilon > 0$, no matter how small, it is decidedly false for $\epsilon = 0$.

