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?