How to show that $\lim\limits_{x \to \infty} f'(x) = 0$ implies $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$?

I was trying to work out a problem I found online. Here is the problem statement:

Let $f(x)$ be continuously differentiable on $(0, \infty)$ and suppose $\lim\limits_{x \to \infty} f'(x) = 0$. Prove that $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$.

The first idea that came to my mind was to show that for all $\epsilon > 0$, we have $|f(x)| < \epsilon|x|$ for sufficiently large $x$. (And I believe I could do this using the fact that $f'(x) \to 0$ as $x \to \infty$.)

However, I was wondering if there was a different (and nicer or cleverer) way. Here's an idea I had in mind:

If $f$ is bounded, then $\frac{f(x)}{x}$ clearly goes to zero. If $\lim\limits_{x \to \infty} f(x)$ is either $+\infty$ or $-\infty$, then we can apply l'Hôpital's rule (to get $\lim\limits_{x \to \infty} \frac{f(x)}{x} = \lim\limits_{x \to \infty} \frac{f'(x)}{1} = 0$).

However, I'm not sure what I could do in the remaining case (when $f$ is unbounded but oscillates like crazy). Is there a way to finish the proof from here?

Also, are there other ways of proving the given statement?

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Certainly, a nicely phrased question. – Srivatsan Sep 8 '11 at 19:08
Doesn't applying L'Hopital's rule directly give this result? – Nima Bavari Feb 7 at 21:04

This is an immediate consequence of L'Hopital's rule. For example, below is said L'Hospital's rule, from Rudin's $\:$ Principles of Mathematical Analysis, $\:$ 1976. Note that it requires only that the denominator $\to\infty\:,\:$ not also the numerator. For more see the Monthly papers cited here.

REMARK $\$ L'Hospital's rule (LHR) is essentially a form of the Mean value Theorem (MVT) repacked into a form convenient for limit calculations. One can of course "unpackage" the MVT and apply it directly without any mention of LHR. It's worth emphasizing that doing so does not really avoid L'Hopital's Rule (LHR) since it is precisely the proof of LHR, only specialized to a specfic function. Further, the proof of most special cases isn't much simpler than the proof of the general case of LHR. The raison d'être of the LHR abstraction is that it encapulates such applications of the Mean Value Theorem into a conveniently applicable form, so that one can easily reuse the proof by simply invoking the rule by name, not by value, i.e. not by repeating the whole proof ("inlining" it) every time one applies it!

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Interesting, most L'Hopital explatations require that also the numerator must tend to infinity, I believed it was required - and so does Wikipedia – leonbloy Sep 8 '11 at 19:44
@Leo This more general version is not as well-known as it deserves to be, esp. considering its proof is only slightly more work than the proof of the less general version. One can find further discussion and generalizations in the cited papers. – Bill Dubuque Sep 8 '11 at 19:57
Wow! This really is an immediate consequence of L'Hopital's rule! Thanks for helping me finish my solution! – Alan C Sep 8 '11 at 23:16
@BillDubuque It the first statement of the original problem changes to $lim f'(x) = A$, does it imply $lim f(x)/x = A$? – Taozi Mar 24 at 6:28

Here's a proof that avoids L'Hôspital's rule (though you can certainly use it, as noted by Bill; basically, the condition on $f'(x)$ ensures that the function cannot oscillate like crazy).

Let $\epsilon\gt 0$. We want to show that there exists $N\gt 0$ such that if $x\gt N$, then $$\left|\frac{f(x)}{x}\right| \lt\epsilon.$$

We know that there exists $M\gt 0$ such that $|f'(x)|\lt \frac{\epsilon}{2}$ for all $x\gt M$. By the Mean Value Theorem, if $x\gt M$, there exists $c\in (M,x)$ such that $f(x)-f(M)=f'(c)(x-M)$, so $$|f(x)-f(M)| = |f'(c)|(x-M) \lt \left(\frac{\epsilon}{2}\right) x.$$ That is, for all $x\gt M$, $$\left|\frac{f(x)}{x} - \frac{f(M)}{x}\right| \lt\frac{\epsilon}{2}.$$

As $x\to\infty$, we know $\frac{f(M)}{x}\to 0$ (since $f(M)$ is fixed). So there exists $N\gt M$ such that if $x\gt N$, then $\left|\frac{f(M)}{x}\right|\lt \frac{\epsilon}{2}$. Thus, if $x\gt N$, then $$\left|\frac{f(x)}{x}\right| \leq \left|\frac{f(x)-f(M)}{x}\right| + \left|\frac{f(M)}{x}\right| \lt \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon,$$ which is what we needed to prove. QED

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It's worth emphasizing that the above proof doesn't really avoid L'Hopital's Rule (LHR) since it is precisely the proof of LHR, only specialized to this function. Further, the proof of this special case isn't much simpler than the proof of the general case of LHR. The raison d'être of the LHR abstraction is that it encapulates such applications of the Mean Value Theorem into a conveniently applicable form, so that one can easily reuse the proof by simply invoking the rule by name, not by value, i.e. not by repeating the whole proof ("inlining" it) every time one applies it. – Bill Dubuque Sep 9 '11 at 0:41
A generalization of this proof gives L'Hopital's rule. Cauchy would be proud of you! – Nima Bavari Feb 7 at 21:08

You can do a proof by contradiction. If ${\displaystyle \lim_{x \rightarrow \infty} {f(x) \over x}}$ were not zero, then there is an $\epsilon > 0$ such that there are $x_1 < x_2 < ... \rightarrow \infty$ such that ${\displaystyle\bigg|{f(x_n) \over x_n}\bigg| > \epsilon}$. Then for $a \leq x_n$ one has $$|f(x_n) - f(a)| \geq |f(x_n)| - |f(a)|$$ $$\geq \epsilon x_n - f(a)$$ By the mean value theorem, there is a $y_n$ between $a$ and $x_n$ such that $$|f'(y_n)| = {|f(x_n) - f(a)| \over x_n - a}$$ $$\geq {\epsilon x_n - f(a) \over x_n - a}$$ Letting $n$ go to infinity this means for sufficiently large $n$ you have $$|f'(y_n)| > {\epsilon \over 2}$$ Since each $y_n \geq a$ and $a$ is arbitrary, $f'(y)$ can't go to zero as $y$ goes to infinity.

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but this is a contradiction only if $y_n\to \infty$. Why you can claim this? Also, when "Letting $n$ go to infinity..." I think $f'(y_n)$ must be go to some thing. That's the problem. For example is for $n$ large, probably $x_n>1$, so, $1\leq y_n<x_n$, there is no reason to prevent $y_n=2$ for all $n$. – leo Sep 8 '11 at 19:51
You can replace 1 with any fixed $a$ so that each $y_n \geq a$. I corrected my answer accordingly, thanks for pointing that out. – Zarrax Sep 8 '11 at 20:31

I'd start like this: for $x\geq 1$ we have $${f(x)\over x}={f(1)\over x}+{1\over x}\int_1^x f^\prime(y)\, dy.$$

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This is really nice! Thanks! – Alan C Sep 8 '11 at 23:18
You're welcome, though it is only a hint and not a full solution. Considering $f(x)/x$ as a kind of average of $f^\prime(y)$ leads to a proof. – Byron Schmuland Sep 8 '11 at 23:36
@Byron Schmuland: your hint is correct not only for $x\geq 1$, for all $x$ positive. – leo Sep 9 '11 at 6:03
Sorry being slow, but I can't think of a way to finish off this proof without doing an epsilon - delta argument similar to Arturo Magidin's using the mean value theorem for integrals. Is there an ending to this proof as elegant as the beginning? – EuYu Oct 21 '11 at 5:32
@EuYu Sorry, I can't think of a better way to finish off the problem than Arturo's solution. – Byron Schmuland Oct 21 '11 at 15:15