Let $f : \mathbb R\to\mathbb R$ and $\ell\in\mathbb R$, and assume that, for every increasing sequence $(\omega_n)\in\mathbb R^\mathbb N$ such that $\lim_{n\to\infty}\omega_n= \infty$, one has $\lim_{n\to\infty} f(\omega_n) = \ell$. Is it necessarily true then, that $\lim_{x\to\infty} f(x) = \ell$ ?
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1Yes, it is true. – markvs Nov 05 '21 at 14:55
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2Well, if the limit is not $\ell$ then for some positive $\epsilon$ it must be the case that there are arbitrarily large values $x_n$ for which $|f(x_n)-\ell|>\epsilon$. use that. – lulu Nov 05 '21 at 14:56
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@lulu I see, in that case one can construct an increasing sequence $(x_n)$ that is unbounded (and thus, will go to infinity) and satisfies $|f(x_n) - \ell| > \epsilon$ for all $n$... Thank you! – KCJV Nov 05 '21 at 15:04
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Yep, that's precisely the idea. – lulu Nov 05 '21 at 15:06
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1If you reject ${\bf CC}(\Bbb R)$, the axiom of countable choice for subsets of $\mathbb R$, this can be false. This answer by Asaf Karagila shows the idea. – Mike Earnest Nov 05 '21 at 15:26