TITLE: Why must delta be defined in terms of epsilon only?
QUESTION [6 upvotes]: I'm a mathematics tutor trying to get a better understanding of the theory of epsilon-delta limit proofs.
I can prove linear and constant epsilon delta proofs easily, and I understand the proof form and definitions, but nonlinear ones stump me.
For example, take Problem 4., Page 3 here:
$$
\lim_{x \rightarrow 2} {x^2 + x - 2} = 4
$$
The paper works through it as follows:
\begin{align}
|f(x)-L| < \epsilon &\implies |(x^2+x-2) - 4| < \epsilon \\
&\implies |(x^2+x-6)| < \epsilon \\
&\implies|x+3||x-2| < \epsilon \\
&\implies |x-2| < \frac{\epsilon}{|x+3|}
\end{align}
Now, this is the point where I would add
$$
\text{let} \ \delta = \frac{\epsilon}{|x+3|}
$$
However, no source I've seen does that. This is where I get a little confused. The epsilon-delta definition means that the expression
$$
\lim_{x\rightarrow c }f(x) = L
$$
is equivalent to
$$
\forall\epsilon >0 \ \exists\delta>0 \ s.t. 0 < |x-c| < \delta \implies |f(x) - L|<\epsilon.
$$
Now, the definition has no qualifications for $x$. $x$ is just the input of the function. However, the paper goes on to say "In general delta must be
in terms of epsilon only, without any extra variables." Why? The proof, when done "forward", goes through just fine for $\epsilon: \epsilon(\delta, x)$. And intuitively, this makes sense: for some parts of $f(x)$, the limits on epsilon will be different. However, every source I've seen finds limits on $|x+3|$ in some region and uses a constant in its place. Why must this be done? Why not leave $|x+3|$ in the denominator and be done with it?

REPLY [2 votes]: Many answers have emphasized what the definition of limit you've given means, i.e. that "$ \forall \epsilon \; \exists \delta$ ..." means that delta must be determinable from epsilon, but it seems none have answered the primary question of why this must be the definition to appropriately capture the notion of a limit. I'll try to address this. First, to established our intuition, the original definition ensures that no matter how small a region around $L$ we choose, we can find some neighborhood of $c$ on which $f(x)$ is always within that small region.
Let's consider what the definition according to your suggested interpretation might look like; that is, for a function $f:U \to \mathbb{R}$ ($U \subset \mathbb{R}$ being the domain of $f$) suppose we define 
$$\lim_{x \to c} f(x) = L \iff \left( \forall \epsilon>0 \; \exists\,\delta_\epsilon:U \backslash \{c\} \to \mathbb{R}_+ \text{ s.t. } |x-c| < \delta_\epsilon(x) \implies |f(x)-L| < \epsilon \right).$$
There's nothing self-contradictory about this proposed definition at the outset-- one can certainly write this down and choose to work with it. However, it utterly fails to capture what a limit should mean (or much of anything, for that matter), the primary reason being that $|x-c| < \delta_\epsilon(x)$ need not be satisfied by any (let alone all) $x$ sufficiently close to $c$ for arbitrary functions $\delta_\epsilon$. A limit should tell us what $f$ is doing everywhere around $c$, and the condition stipulated by this definition says nothing about those $x$ near $c$ that don't satisfy $|x-c| < \delta_\epsilon(x)$. What's more, this definition makes limits (very) nonunique.
To see these objections more clearly, consider the prototypical step function
$$f(x) =
\begin{cases}
1,  & x \geq 0 \\
0, & x < 0
\end{cases}$$
I expect we can agree that the limit $\lim_{x \to 0} f(x)$ should not exist, i.e. there is no one number that $f$ tends to "approach" as $x \to 0$. However, given $\epsilon > 0$, consider the function 
$$\delta_\epsilon(x) =
\begin{cases}
2|x|,  & x > 0 \\
|x|/2, & x < 0
\end{cases}$$
Then the statement that $|x-0| < \delta_\epsilon(x) \implies |f(x)-1| < \epsilon$ is true, since $|x-0| < \delta_\epsilon(x)$ is only satisfied for $x>0$, where $f$ is identically equal to $1$. That is, according to our proposed definition, $\lim_{x \to 0} f(x) = 1$! A very simple modification also demonstrates that $\lim_{x \to 0} f(x) = 0$. But the situation is even (much) worse than this! By choosing $\delta_\epsilon$ such that $|x-c| < \delta_\epsilon(x)$ is never satisfied (say $\delta_\epsilon(x) = |x-c|/2$), the definition is shown to be vacuously satisfied for any $L$ whatsoever!
We're therefore forced to the following conclusion: allowing $\delta$ to depend on $x$ as well simply breaks the definition's meaning entirely. Placing the necessary restrictions on the functions $\delta_\epsilon$ to make the above definition reasonable will ultimately give you the standard definition.