For a generic (real) function $g(x)$ and $x \in \mathbb{R}_{\geq 0}$, how can I find the largest $f(x)$ for which $f(x)/g(x)\to 0$ as $x \to \infty$? This question is not for any reason beyond personal curiosity.
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2$\begingroup$ Can you define precisely what you mean by the largest $f$? The largest usual involves an order relation. $\endgroup$– mathcounterexamples.netMay 19 at 17:48
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$\begingroup$ As long as $f(x)$ does not create a $\dfrac{\infty}{\infty}, 0 \cdot \infty$ or other similar problem, the "largeness" of $f$ is irrelevant. That said, the wording is very poor. $\endgroup$– algevristisMay 19 at 17:54
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$\begingroup$ Let $a>0$ and $\mathcal{H}$ be the set of all real functions $h(x)$ such that $h(x)/g(x) \to 0$ as $x \to \infty$. By the largest $f(x)$, I mean the $f(x)$ such that $h(x)\leq f(x)$, for all $x \geq a$, for all $h \in \mathcal{H}$. @mathcounterexamples.net $\endgroup$– asdfglillMay 19 at 17:56
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2$\begingroup$ If $f, g \to \infty$ with $f/g \to 0$ then $F= \sqrt{fg}$ satisfies $F/g \to 0$ and $F/f \to \infty$. So for every $f$ there is some “larger” $F$. $\endgroup$– Martin RMay 19 at 18:00
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$\begingroup$ Not sure that downvoting such a question brings benefit for the person who asked the question, nor for the one who downvoted, and not also for the community. $\endgroup$– mathcounterexamples.netMay 19 at 18:03
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1 Answer
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Realistically, there is no largest such $f(x)$. Simply take
$$f(x)=\frac{g(x)}{r(x)}$$
for any $r(x)$ such that
$$\lim_{x\to\infty}r(x)=\infty$$
This then gives
$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\lim_{x\to\infty} \frac{1}{r(x)}=0$$
So your question is basically: Is there a slowest function growing to infinity. There is not. To see why, simply note that for any function which grows slower than $\log(x)$ in the sense that
$$\lim_{x\to\infty}\frac{r(x)}{\log(x)}=0$$
then $r(r(x))$ grows slower than $r(x)$.
