Thoughts about the use of $\epsilon$-$\delta$ language in proofs involving limits The limit of a function $f: \mathbb{R} \to \mathbb{R}$ at $x_{0}$ is defined as follows:

$\lim_{x \to x_{0}} f\left(x\right) = L \iff \forall \epsilon > 0, \exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon$.

In proofs of limits using the definition, we typically fix an arbitrary constant $\epsilon > 0$, and as long as we are able to prove
\begin{equation}
\exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon,
\end{equation}
we can conclude that $\lim_{x\to x_{0}}f\left(x\right) = L$. The key is to construct such a $\delta$ so that the above relation holds. However, as I saw in some textbooks, for an arbitrary $\epsilon > 0$, if we are able to find a $\delta > 0$ such that
\begin{equation}
\forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < g\left(\epsilon\right),
\end{equation}
where $g\left(\epsilon\right)$ is a term containing $\epsilon$ ($g\left(\epsilon\right) = 3\cdot\epsilon$, for example),
then we can still claim that $\lim_{x\to x_{0}}f\left(x\right) = L$.
I am wondering if there exists a simple proof for using the general $g\left(\epsilon\right)$ in proofs of limits?
 A: If you have $|A|<3\epsilon$ for any $\epsilon>0$, you can conclude that $|A|<\epsilon$ for any $\epsilon>0$, which implies that $|A|=0$.
In general, if you have the estimates of the form $|A|<g(\epsilon)$ for all $\epsilon>0$, where $\displaystyle\lim_{y\to 0+}g(y)=0$, you can conclude that $|A|=0$.

Suppose you have shown that
$$
\forall \epsilon\ \exists \delta\ \forall x \ \ |x-x_0|<\delta\Rightarrow |f(x)-L|<g(\epsilon)\tag{1}
$$
where $\displaystyle \lim_{y\to 0+}g(y)=0$,
you can then show that
$$
\forall \epsilon\ \exists \delta\ \forall x \ \ |x-x_0|<\delta\Rightarrow |f(x)-L|<\epsilon\tag{2}
$$
Proof.
Let $\epsilon>0$. Since $\displaystyle \lim_{y\to 0+}g(y)=0$, there exists $r>0$ such that $g(r)<\epsilon$. By (1), there exists $\delta>0$ such that
$$
|x-x_0|<\delta\Rightarrow |f(x)-L|<g(r)<\epsilon.
$$
Now you have (2).
A: As long as $g(\epsilon)$ goes infinitely close to $0$ while staying positive, as $\epsilon \to 0$, i.e. $\lim_{\epsilon\to 0} g(\epsilon) = 0$ and $g(\epsilon) > 0$, any such $g$ will do because
$\forall \zeta > 0,\ \exists \gamma > 0, \ 0 < \epsilon < \gamma \implies g(\epsilon) < \zeta$
So, if using an $\epsilon-\delta$ technique we conclude that $\exists$ a suitable $\delta$ such that $|f(x) - L| < g(\epsilon)$, then $\forall \zeta > 0$, we choose $\epsilon < \gamma$ and $|f(x) - L| < g(\epsilon) < \zeta$ whenever $0 < |x-x_0| < \delta$.
A: To formalize @DaniloGregorinAfonso's idea, you may set $\epsilon' = g(\epsilon)$, so as to cheer up yourself for the moment, giving a hope that you can "clean up the RHS".
$$\forall \epsilon > 0, \exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon'$$
Assume that $\epsilon = g^{-1}(\epsilon')$.  Note that $\delta$ is a function $\epsilon$, so you may write $\delta = \delta(\epsilon) = \delta(g^{-1}(\epsilon'))$.  You may find the notation $g^{-1}$ problematic, say, in case that $g(\epsilon) = \begin{cases}\epsilon (\sin(1/\epsilon) + 2021/2020) & \epsilon \ne 0 \\ 0 & \epsilon = 0. \end{cases}$  In such case, you may find a monotone increasing upper bound $h(\epsilon) \ge g(\epsilon)$ such that $h(\epsilon) \to 0$ as $\epsilon \to 0$, and do that same thing to get $\delta(\epsilon) = \delta(h^{-1}(\epsilon'))$.  Now you get a function $\delta$ which depends on the new $\epsilon'$.
$$\forall \epsilon > 0, \exists \delta(h^{-1}(\epsilon')) > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon'$$
Finally replace $\epsilon$ in "$\forall \epsilon > 0$" on the left-hand side by $\epsilon'$.
It remains to show that such monotone increasing upper bound $h$ for $g$ exists, but we may construct it like $h(\epsilon) = \sup\limits_{t \in (0,\epsilon)} |g(\epsilon)|$.  It's clear that $h$ is monotone increasing.  It's easy to verify that $h\searrow0$ as $\epsilon \searrow 0$.

 $$\forall \epsilon > 0, \exists \delta > 0, \forall t \in (0,\delta), |g(t)| < \epsilon \\ \implies \sup_{t \in (0,\delta)} |g(t)| < \epsilon \\ \implies \forall t \in (0,\delta), h(t) = \sup_{s \in (0,t)} |g(s)| \le \sup_{s \in (0,\delta)} |g(s)| < \epsilon$$

Add an extra term $\epsilon$ to $h$ guarantees that $h$ to be strictly increasing, i.e. the existence of $h^{-1}$, completing the proof.
