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The sequential criterion for functional limits states that given a function $f:A \rightarrow \mathbb{R}$ and a limit point $c$ of $A$, the following two statements are equivalent:

  1. $\lim_{x \rightarrow c} f(x) = L$

  2. For all sequences $(x_n) \subset A$ satisfying $x_n \neq c$ and $(x_n) \rightarrow c$, it follows that $f(x_n) \rightarrow L$.

I was wondering, am I correct in saying the following is also true?

Given a function $f:A \rightarrow \mathbb{R}$ and a limit point $c$ of $A$, the following two statements are equivalent:

  1. $\lim_{x \rightarrow c} f(x) $ does not exist.

  2. There exists a sequence $(x_n) \in A$ satisfying $x_n \neq c$ and $(x_n) \rightarrow c$, but $\lim_{n \rightarrow \infty} (f(x_n))$ does not exist.

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  • $\begingroup$ Yes, that's correct. $\endgroup$ Apr 19, 2014 at 17:12
  • $\begingroup$ Thanks. Any hints on how I can prove it? $\endgroup$
    – user144283
    Apr 19, 2014 at 17:13
  • $\begingroup$ 2. implies 1. should be straightforward. Assume 1. Pick $x_n\rightarrow c$ appropriately. If $(|f(x_n)|)$ is not bounded, we're ok. Otherwise, pick a convergent subsequence of $(f(x_n))$ with limit $a$, say. Since $\lim_{x\rightarrow c}f(x)\ne a$, you can find an admissable sequence $(y_n)$ with limit $c$ and such that $(f(y_n))$ is bounded away from $a$. Consider the sequence $x_1,y_1,x_2,y_2,\ldots$. $\endgroup$ Apr 19, 2014 at 17:19
  • $\begingroup$ @DavidMitra But, how does $x_1,y_1,...$ does the job? $\endgroup$ Aug 27, 2023 at 8:30
  • $\begingroup$ @ThomasFinley There is a subsequence converging to $a$. The sequence itself can't converge to $a$ (which it would have to if it converged), since the $f(y_n)$ are bounded away from $a$. $\endgroup$ Aug 27, 2023 at 10:46

1 Answer 1

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You haven't specified a nature of $A$. If $A\subset \mathbb R$, the equivalence of 1 and 2 holds. It also holds if the domain of $f$ is a metric space (or a first-countable topological space).

Anyway, the comment by David Mitra is essentially the solution. I give a slightly different version of "(1) implies (2)". First, take any sequence $(x_n)$ in $A\setminus \{c\} $ converging to $c$. If $\lim_{n\to\infty} f(x_n)$ does not exist, we are done. Otherwise, let $L=\lim_{n\to\infty} f(x_n)$. Since $\lim_{x\to c} f(x)$ does not exist, the negation of the definition of a limit says: there exists $\epsilon>0$ such that for any $\delta>0$ there is $x\in A\setminus \{c\}$ with $|f(x)-L|\ge \epsilon$. Apply this fact with $\delta=1/n$, and let $y_n=x$. The sequence $x_1,y_1,x_2,y_2,\dots$ does the job.

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