Lately, I've been interested in the many ways one could make the intuitive idea of the "limit of a function" mathematically rigorous. Among the various definitions I've come up with and tested, one stood out as particularly appealing:
Definition: if $f$ is defined around $a$, then we say that the limit of $f(x)$ as $x$ approaches $a$ is $L$ if and only if for every pair of positive numbers $(\varepsilon_1,\varepsilon_2)$, there exists a pair of positive numbers $(\delta_1,\delta_2)$ such that for every $x$ in the domain of $f$, if $a-\delta_1<x<a+\delta_2$, then $L-\varepsilon_1<f(x)<L+\varepsilon_2$.
This definition is clearly very similar to the ordinary $(\varepsilon,\delta)$ definition, the only difference being that we use different $\varepsilon$s and $\delta$s for the "upper" and "lower" error tolerances. I originally believed that this definition generalized the $(\varepsilon,\delta)$ one, since we can still make $\varepsilon_1$ and $\varepsilon_2$ arbitrarily small and it doesn't constrain us to use open intervals centered at $a$ and $L$. However, upon further reflection, it seems like these definitions are actually equivalent. I wrote a proof to justify this conclusion, and would love it if someone took the time to review it!
Note: for clarity, when I write $\lim_{x\to a}f(x)=L$, the expression should be interpreted in the context of the ordinary $(\varepsilon,\delta)$ definition. Also, for brevity's sake, the property in the definition I gave will be denoted by $P(f)$.
Proof: Let's first prove that $P(f)$ implies $\lim_{x\to a}f(x)=L$. Given $\varepsilon>0$, we need to find a $\delta>0$ such that for every $x\in\text{dom}[f]$, if $0<|x-a|<\delta$, then $|f(x)-L|<\varepsilon$. We know that $P(f)$ is true, so for every pair of positive numbers $(\varepsilon_1,\varepsilon_2)$, there exists a pair of positive numbers $(\delta_1,\delta_2)$ such that for every $x\in\text{dom}[f]$, if $a-\delta_1<x<a+\delta_2$, then $L-\varepsilon_1<f(x)<L+\varepsilon_2$. In particular, this is true if $\varepsilon_1=\varepsilon_2=\varepsilon$, so there exists a pair of positive numbers $(\delta_1,\delta_2)$ such that if $a-\delta_1<x<a+\delta_2$, then $L-\varepsilon<f(x)<L+\varepsilon$.
Let $\delta=\min\{\delta_1,\delta_2\}$. Then for any $x\in\text{dom}[f]$, if $0<|x-a|<\delta$, then $0<|x-a|<\delta_1$ and $0<|x-a|<\delta_2$, so
\begin{align*} a-\delta_1 &< x<a+\delta_1\\ a-\delta_2 &< x<a+\delta_2 \end{align*}
are both true. These imply that $a-\delta_1 < x<a+\delta_2$, so $L-\varepsilon<f(x)<L+\varepsilon$, and thus $|f(x)-L|<\varepsilon$.
$\varepsilon$ was given, so the same argument applies to every positive real number $\varepsilon$. Thus, $\lim_{x\to a}f(x)=L$.
Now let's prove the converse. Given a pair of positive numbers $(\varepsilon_1,\varepsilon_2)$, we need to find a pair of positive numbers $(\delta_1,\delta_2)$ such that for every $x\in\text{dom}[f]$, if $a-\delta_1<x<a+\delta_2$, then $L-\varepsilon_1<f(x)<L+\varepsilon_2$. We know that $\lim_{x\to a}f(x)=L$ is true, so for every $\varepsilon>0$, there is a $\delta>0$ such that for every $x\in\text{dom}[f]$, if $0<|x-a|<\delta$, then $|f(x)-L|<\varepsilon$. In particular, this will be true if $\varepsilon=\min\{\varepsilon_1,\varepsilon_2\}$ (the minimum of two positive numbers is also positive), so there is a $\delta>0$ so that if $0<|x-a|<\delta$, then $|f(x)-L|<\min\{\varepsilon_1,\varepsilon_2\}$. The latter inequality implies that $|f(x)-L|<\varepsilon_1$ and $|f(x)-L|<\varepsilon_2$, so
\begin{align*} L-\varepsilon_1 &< f(x)<L+\varepsilon_1\\ L-\varepsilon_2 &< f(x)<L+\varepsilon_2 \end{align*}
are both true. Thus, $L-\varepsilon_1<f(x)<L+\varepsilon_2$.
This holds for $0<|x-a|<\delta$, or $a-\delta<x<a+\delta$, so we can take $\delta_1=\delta_2=\delta$. We have thus shown that there exists a pair of positive numbers $(\delta_1,\delta_2)$ such that for every $x\in\text{dom}[f]$, if $a-\delta_1<x<a+\delta_2$, then $L-\varepsilon_1<f(x)<a+\varepsilon_2$.
Since $(\varepsilon_1,\varepsilon_2)$ was given, the same argument applies to every pair of positive numbers. Thus, $P(f)$ is true.
Let me know what you think! If you identify any mistakes, please don't hesitate to share them with me.
Edit: as the comments kindly pointed out, my definition should've included the constraint $x\neq a$. I originally thought my definition would imply this, but I now realize I'm mistaken: if $a-\delta_1<x<a-\delta_2\iff-\delta_1<x-a<\delta_2$ and $\delta_1$, $\delta_2$ are both positive, then $x-a=0$ is still a possibility, causing problems like the ones mentioned in the comments. I won't tweak the original post to include this constraint (I feel that this would be an act of dishonesty), but from this point forward, it should be assumed that $x\neq a$ where relevant.