# The largest $f(x)$ for which $f(x)/g(x)\to 0$ as $x \to \infty$ [closed]

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.

• Can you define precisely what you mean by the largest $f$? The largest usual involves an order relation. May 19 at 17:48
• 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. May 19 at 17:54
• 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 May 19 at 17:56
• 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” $F$. May 19 at 18:00
• 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. May 19 at 18:03

## 1 Answer

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 $$r(x)$$.