Let $f$ be a function defined on a subset $S$ of $\mathbb{R}$, let $a$ be a real number that is the limit of some sequence in $S$, and let $L$ be a real number. Let lim $n \rightarrow a $ along $S$ of $f(x) = L $ be defined as: "For every sequence $(x_n)$ in $S$ with limit $a$, we have lim $n \rightarrow \infty $ along $S$ of $f(x_n) = L $."
Using this sequential definition I am trying to prove: lim $n \rightarrow a $ along $S$ of $f(x) = L $ implies:
for each $\epsilon > 0$, there exists $\delta > 0$ such that $x \in S$ and $| x -a |< \delta$ imply $|f(x) - L| < \epsilon$.
Is this valid?
"Let $\epsilon > 0$. Then for some $x_n$, there exists $N$ s.t. $ n > N \implies |f(x_n) - L| < \epsilon$. Let $\delta = | f^{-1}(x_n) - a |$. Then $|x_n - a| < \delta \implies |f(x_n) - L | < \epsilon$."
EDIT in response to Celio's comment.
"Let $\epsilon > 0$. Then for some sequence $x_n$, there exists $N$ s.t. $ n> N \implies |f(x_n) - L| < \epsilon$. Let $\delta =$ min {$| f^{-1}(x_{N+1}) - a |$}. Then $|x_n - a| < \delta \implies |f(x_n) - L | < \epsilon$."
Here I mean take the min of the distance of each preimage of $x_{N+1}$ from $a$. (Not the inverse).
I would really value any additional feedback!
First, the statement "Let $\delta = min\{|f^{-1}(x_{N + 1}) - a|\}"$ is not clear enough. It may be that $f^{-1}(x_{N + 1}) = \emptyset$.
In your proof, there's an assumption that you can take N to be uniform across all sequences in S that converges to a. This is something you have to prove and you have to show that $\delta$ as defined is positive.
Also what you're trying to show is: for all $\epsilon > 0$, there exists a $\delta >0$ such that if $x \in S$ and $|x - a| \leq \delta$ then $|f(x) - L| \leq \epsilon$. But what you end up showing is: "$|x_{n} - a| \leq \delta \implies |f(x_{n}) - L| \leq \epsilon$ for large n, but it's not clear what the sequence $(x_{n})$ is.
Usually this statement is proven with a proof by contradiction. It may be easier to attempt it that way. Assuming that $\lim_{x \to a} f(x) \not = L$, then there exists an $\epsilon > 0$ such that for all $\delta > 0$, there exists an $x \in S$ such that $|f(x) - L| > \epsilon$ and $|x - a| \leq \delta$. Hence, for the sequence $\delta_{n} = 1/n$ for all $n \geq 1$, we get a sequence of $(x_{n}) \subset S$ that converges to a but $\lim_{n \to \infty} f(x_{n}) \not = L$.