Are these two statements on limits equivalent?

Question

For some real valued function $f$, $\lim\limits_{x \to c}f(x) =L$ is equivalent to saying for any real sequence ${x_n}$, if $x_n$ converges to $c$, then $f(x_n)$ converges to $L$.

If this statement is true, I want to prove it using epsilon delta definition. For the latter statement, we know, $\forall \epsilon >0, \exists N \in \mathbb{N}$ so that $\forall n>N$ we have $0<|x_n-c|< \epsilon$. Now this should imply $f(x_n) \to L$ How do I show it?

For delta-epsilon, you don't want $0<|x_n-c|$; it's OK if $x_n=c$ for some (or even all) $n$. $|x_n-c|<\epsilon$ is all you need/want. — Gerry Myerson, Dec 02 '23 at 06:39
@GerryMyerson $c$ itself may or may not be in the domain of $f$ in this case — zaemon_23, Dec 02 '23 at 06:41
@Andrew i found this statement in one of the answers here https://math.stackexchange.com/a/2967382/1215260 — zaemon_23, Dec 02 '23 at 06:42
See also https://math.stackexchange.com/questions/1992204/precise-proof-of-equivalence-of-two-definitions-of-limit-of-a-function — Gerry Myerson, Dec 02 '23 at 06:47

score 0 · Answer 1 · answered Dec 02 '23 at 12:52

Consider: $$ f(x)= \begin{cases} 1 \quad & \text{if $x\neq 0$}\\ 0 \quad & \text{if $x =0$} \end{cases}$$ Consider also $x_n =0$ for all $n$. Then $x_n\to 0$, and $\lim_{x\to 0} f(x) =1$, but $f(x_n) = 0\,\, \forall$ $n$ and hence $f(x_n) \not \to 1$.

As pointed out in the comments, if you keep $c$ outside the range of the sequence, then the equivalence holds as proved in one of the links mentioned. Following is the equivalence:

Suppose $f(x)$ is a real valued function. Then $\lim\limits_{x\to c} f(x) = L$ if and only if for every sequence $\{x_n\}$ with $x_n\neq c$, for all $n$, we have $\lim\limits_{n\to \infty} f(x_n) = L$, whenever $x_n\to c$.

Are these two statements on limits equivalent?

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