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Suppose that $$f\colon (a,\infty) \to \mathbb{R}$$ is differentiable, and that $$\lim_{x\to\infty} f'(x) =0. $$

Show that $$\lim_{x\to\infty} \frac{f(x)}{x} = 0.$$

This can be done easily with L'Hôpital's rule, but it appears as a textbook exercise, in a chapter discussing Rolle's theorem, Mean Value Theorem, etc., before L'Hôpital's rule is introduced.

Teddy
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    Apply MVT to $f(x)-f(T)$ where $x>T$ and $T$ is sch that $|f'(t)| <\epsilon$ for $ t >T$. – Kavi Rama Murthy Dec 22 '20 at 07:31
  • Note that we can have $f'(c)=\frac{f(x)-f(0)}{x}$ for some $c\in(0,x)$.Now use $ c\to \infty\implies x\to \infty$. – PNDas Dec 22 '20 at 07:33
  • @KaviRamaMurthy It's a little tricky. Can you spell it out ? We can say that $\frac{f(x)-f(T)}{x-T} = f'(c_x) < \epsilon$. But how do you get rid of the $T$ and $f(T)$ ? – Teddy Dec 22 '20 at 08:30
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    Does this help: https://math.stackexchange.com/questions/298832/if-lim-x-to-infty-fx-l-then-lim-x-to-infty-frac-fxx? – Hans Lundmark Dec 22 '20 at 08:35
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    @PNDas: that argument does not work because what we need here is the implication $x\to\infty \implies c\to\infty $ which is not guaranteed to hold. – Paramanand Singh Dec 22 '20 at 10:53
  • This is essentially reproducing the proof of L'Hospital's Rule for the "$\text{anything} /\infty$" case. – Paramanand Singh Dec 22 '20 at 10:54
  • @ParamanandSingh,Thank you for your comment. But can't we say $\lim_{c\to \infty}f'(c)=\lim_{c\to\infty}\frac{f(x)}{x}-\frac{f(0)}{x}=\lim_{x\to\infty}\frac{f(x)}{x}-\frac{f(0)}{x}$? Please note that when I was writing the first comment I didn't see that the domain is greater than $a$. Even if the domain is greater than $a$ we can take any $b>a$ in place of $0$. – PNDas Dec 22 '20 at 11:52
  • @PNDas: no we can't say like that. Two reasons : first we can't switch from $c$ to $x$ as $x$ is not a function of $c$. And second $c\to\infty $ does not imply $x\to\infty $. – Paramanand Singh Dec 22 '20 at 13:07
  • @ParamanandSingh, Actually I was saying since $0<c<x$ then $c\to\infty\implies x\to \infty$ but I found my mistake. Thank you. – PNDas Dec 22 '20 at 13:12
  • @PNDas: all we know for sure is that $0<c<x$ and $x\to\infty $. We don't know if $c\to\infty $. – Paramanand Singh Dec 22 '20 at 16:17

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Let $x >T$. $\frac {f(x)} x=\frac {f(x)-f(T)} x +\frac {f(T)} x=\frac {x-T} x \frac {f(x)-f(T)} {x-T} +\frac {f(T)} x$By MVT this gives $\frac {f(x)} x=\frac {x-T} x f'(s) +\frac {f(T)} x$ for some $s \in (T,x)$. Can you finish?

Kavi Rama Murthy
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