How to show that limx→∞f′(x)=0\lim\limits_{x \to \infty} f'(x) = 0 implies limx→∞f(x)x=0\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,) and suppose limxf(x)=0. Prove that limxf(x)x=0.


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

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 f(x)x clearly goes to zero. If limxf(x) is either + or , then we can apply l'Hôpital's rule (to get limxf(x)x=limxf(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?


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 , 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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Source : Link , Question Author : Alan C , Answer Author : Bill Dubuque

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