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

Question

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 $\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0$.

(source: http://www.math.vt.edu/people/plinnell/Vtregional/E79/index.html)

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?

Answer 1

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!

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

let $f:\left[ a,+\infty \right[ \longrightarrow \mathbb{R} $ differentiable suppose that $\displaystyle \lim_{x\rightarrow +\infty} f'(x)=l \in \mathbb{R}$ prove that $\displaystyle \lim_{x\rightarrow +\infty} \dfrac{f(x)}{x}=l $

we will begin with the mean value theorem : for $ x > \max(0,a) \ \exists c\in \left]a,x\right[ $ such that $$f(x)-f(a)=f'(c)(x-a)\implies \dfrac{\dfrac{f(x)}{x}-\dfrac{f(a)}{x}}{1-\dfrac{a}{x}}=f'(c)$$ and because as $c\rightarrow +\infty$ , $x\rightarrow +\infty$ then we have: $$\lim_{c\rightarrow +\infty} f'(c)=\lim_{c\rightarrow +\infty }\dfrac{\dfrac{f(x)}{x}-\dfrac{f(a)}{x}}{1-\dfrac{a}{x}}\implies \lim_{c\rightarrow +\infty}\dfrac{\dfrac{f(x)}{x}-\dfrac{f(a)}{x}}{1-\dfrac{a}{x}}=l\implies \lim_{x\rightarrow +\infty}\dfrac{\dfrac{f(x)}{x}-\dfrac{f(a)}{x}}{1-\dfrac{a}{x}}=l$$ thus we have the desired result: $$\lim_{x\rightarrow +\infty} \dfrac{f(x)}{x}=l$$