I came up with an alternate definition of limit, which I would like to verify it is equivalent to the usual one.
A sequence $(a_n)$ has a limit $L$ if, for any $a<L<b$, there are only finitely many $n$ with $a_n<a$ and only finitely many $n$ with $b<a_n$.
In other words, a sequence $(a_n)$ has a limit $L$ if, for any $M\ne L$, there are only finitely many elements of the multiset $\{a_n:n\in\Bbb N\}$ on the "opposite side" of $M$ from $L$.
(Remember that a "multiset" is like a set but it's allowed to have more than one of a single element. For example, the multisets $\{1\}$ and $\{1,1\}$ are different.)
I believe it's equivalent because it implies that any neighborhood of $L$ has cofinitely many elements of the multiset $\{a_n:n\in\Bbb N\}$, meaning there's an $N$ such that for all $m>N$, $a_n$ is in the neighborhood. Is this correct?
(I like the second one better because it seems simpler and more geometric. It doesn't have many quantifiers alternations or any advanced definitions — except for multisets, which seems like a natural idea that everyone should know about anyway.)
Proof of equivalence:
Suppose $\{a_n\}$ possesses the said property. Pick $\varepsilon>0$ and let $ a=L-\varepsilon, b=L+\varepsilon$. As $A:=\{n\in\mathbb{N}:a_n<a\lor a_n>b\}$ is finite define $N$ to be $\max(A)+1$ and we have $a<a_n<b\;\forall\;n\ge N.$ Therefore $n\ge N\implies |a_n-L|<\varepsilon$.
Suppose $\{a_n\}$ converges to $L$. Pick $a<L<b$. Define $A:=\{n\in\mathbb{N}:a_n<a\}$. Let $\varepsilon=L-a$. $\exists N: n\ge N\implies |a_n-L|<\varepsilon\implies a_n>L-\varepsilon=a$. So $A$ is a finite set. Similarly $\{n\in\mathbb{N}:a_n>b\}$ is finite.