It is a bit unconventional but you can do this (and it is even quite useful!). However, you need to have a condition on how to choose epsilon to make it work
Here is a $\delta$-$\epsilon$ definition of continuity of a function $f$ in $x_0$:
$\forall \delta>0, \exists \epsilon = \epsilon(\delta) > 0$ such that
- $\forall x$ with $|x - x_0| \le \delta$ we have $|f(x) - f(x_0)| \le \epsilon(\delta)$
- $\inf_{\delta > 0} \epsilon(\delta) = 0$
It is best to think of the $\forall \delta >0 \exists \epsilon(\delta) > 0$ part as the existence of a function $\epsilon(\delta)$ with certain properties. The second property, ensures that the above counterexamples do not work.
By replacing $\epsilon(\delta)$ by $\epsilon'(\delta) = \inf_{\delta' \ge \delta} \epsilon(\delta)$ we can w.l.o.g. assume that
- if $\delta_1 \le \delta_2$ we have $\epsilon(\delta_1) \le \epsilon(\delta_2)$
This makes it much clearer what is going on, and makes it an easy exercise to prove that the $\delta$-$\epsilon$ definition is equivalent to the $\epsilon$-$\delta$ definition (see below). The function $\epsilon(\delta)$ tells you not only that $f(x) \to f(x_0)$ as $x \to x_0$ but how fast this is happening, i.e. $\epsilon(\delta)$ controls how a deviation of $x$ from $x_0$ results in a (maximal) deviation of $f(x)$ from $f(x_0)$. The $\delta$-$\epsilon$ definition makes many standard proofs or explicit computations much more natural without "magic" choices like "choose $\delta = \epsilon^{100}/\pi^2$" often found in proofs using the $\epsilon$-$\delta$ definition. E.g. to show that the product of two functions is continuous suppose that $\epsilon_f(\delta)$ and $\epsilon_g(\delta)$ control the convergence of $f(x) \to f(x_0)$ and $g(x) \to g(x_0)$. Then for $|x - x_0| < \delta \le \delta_0$ (which is all that matters) we have
$|f(x)g(x) - f(x_0)g(x_0)|
\le |f(x) - f(x_0)||g(x_0)| + |g(x) - g(x_0)||f(x)| \le |g(x_0)|\epsilon_f(\delta) + (|f(x_0)| + \epsilon_f(\delta_0))\epsilon_g(\delta)$
it is trivial that $\epsilon_{fg}(\delta) := |g(x_0)|\epsilon_f(\delta) + (|f(x_0)| + \epsilon(\delta_0))\epsilon_g(\delta)$ satisfies property 2 and 3 and so controls the convergence of $fg(x) \to fg(x_0)$, showing that $fg$ is continuous in $x_0$.
In fact while the $\delta$-$\epsilon$ is unconventional we often use it when we want stronger notions of continuity and stronger control over the convergence of $f(x) \to f(x_0)$
A function f is Lipschitz continuous in $x_0$ if
1. $\forall x$ with $|x - x_0| \le \delta$ we have $|f(x) - f(x_0)| \le C\delta$ for some constant $C > 0$.
Clearly property 2 and 3 are now trivially fulfilled. Often we want a $C$ to be uniform for the whole domain of $f$. If such a $C$ exists we call $f$ Lipschitz continuous.
likewise:
A function is $\alpha$ Hölder continuous for $0 < \alpha$ if
1. $\forall x$ with $|x - x_0| \le \delta$ we have $|f(x) - f(x_0)| \le C\delta^\alpha$ for some constant $C > 0$.
Again property 2 and 3 are trivially fulfilled.
======
Proof that $\delta$-$\epsilon$ continuous $\iff$ $\epsilon$-$\delta$ continuous.
$\implies$:
Choose $\epsilon_0 > 0$. By property 2, $\exists \delta_0$ such that $\epsilon(\delta_0) \le \epsilon_0$ hence $\forall x$ with $|x - x_0| \le \delta_0$ we have $|f(x) - f(x_0)| \le \epsilon_0$.
$\Leftarrow$
By the $\epsilon$-$\delta$ definition, there exists a function $\delta(\epsilon)$ such that
$\forall x$ with $|x - x_0| \le \delta(\epsilon)$ we have $|f(x) - f(x_0)| \le \epsilon$
For $\delta_0 > 0$, let
$\epsilon(\delta_0) = \inf \{\epsilon >0, 0<\delta_0 \le \delta(\epsilon)\}$ [1].
Fix $\epsilon_0 > 0$ and assume $\delta_0 \le \delta(\epsilon_0)$. Then $\forall x$ with $|x- x_0| \le \delta_0\le \delta(\epsilon_0)$ we have $|f(x) - f(x_0)| \le \epsilon_0$. Now reversing points of view and taking any $\epsilon_0$ with $\delta(\epsilon_0) \ge \delta_0$ we see that on taking inf's we get $|f(x) - f(x_0)| \le \epsilon(\delta_0)$ as in property 1. We have chosen $\epsilon$ non decreasing as function of $\delta_0$ as in property 3 by construction. In particular for $\delta_0 \le \delta(\epsilon_0)$ we have $\epsilon(\delta_0) \le \epsilon(\delta(\epsilon_0))\le \epsilon_0$, so $0\le \inf_{\delta_0 > 0} \epsilon(\delta_0) \le \inf_{\epsilon_0 > 0} \epsilon(\delta(\epsilon_0)) \le \inf_{\epsilon_0 >0} \epsilon_0 = 0$ proving propery 2.
[1] We strictly speaking have to assume that $0 < \delta_0 \le \delta(1)$ say, to make sure that the inf is over a non empty set, but that does not matter, we only need control functions in a neighborhood of $\delta_0 = 0$.