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I found a visual proof of the L'Hôpital's rule by Giorgio Goldoni which uses an additional hypothesis: g'(x) (i.e. the function which is at the denominator) cannot change its sign.

My question is this: is this additional hypothesis reductive? Does it exist a function whose first derivative changes sign infinitely many times, but for which the De l'Hospital rule works?

2 Answers 2

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If the denominator changes sign infinitely many times as $x\to a$, then by Darboux' theorem it will be $0$ at points arbitarily close to $a$. That means \lim_{x\to a} \frac{f'}{g'} fails to exist at all, because it can only exist if there is a set $A\ni a$ such that $A$ is open in the domain of $\frac{f}{g}$, and \frac{f'}{g'} is defined everywhere on $A\setminus\{a\}$.

Also, if g' changes sign infinitely often, then so does g'' if it exists (and so forth by induction), so no matter what order of L'Hôpital we use, it will fail to work. Thus, higher-order L'Hôpital cannot work for such functions either.

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    Very good, thank you.2011-11-06
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Update: As it turned out, my answer below isn't correct because $\frac{f'(x)}{g'(x)}$ fraction cannot be simply reduced, because it's not defined in points where $g(x)=0$ (see comments). However, I am not removing the post, because it's can serve as a demonstration how one can make a mistake solving problems like this.


Yes, there do exist such example, but it doesn't mean that the $g'(x)\not=0$ condition is reductive - it's a sufficient condition, but not necessary:

Consider functions $f(x)=-x*sin(\frac{1}{x})$ and $g(x)=2*f(x)$ and assume $a=0$.

We have $\lim_{x \to 0}f(x)=0$ and $\lim_{x \to 0}g(x)=0$

enter image description here

From one hand, obviously, $\lim_{x \to 0}\frac{f(x)}{g(x)}=\lim_{x \to 0}\frac{f(x)}{2*f(x)}=\frac{1}{2}$

Now, $f'(x)=\frac{cos(\frac{1}{x})}{x}-sin(\frac{1}{x})$ and $g'(x)=2*\frac{cos(\frac{1}{x})}{x}-2*sin(\frac{1}{x})$

And here we also have $\lim_{x \to 0}\frac{f'(x)}{g'(x)}=\lim_{x \to 0}\frac{\frac{cos(\frac{1}{x})}{x}-sin(\frac{1}{x})}{2*\frac{cos(\frac{1}{x})}{x}-2*sin(\frac{1}{x})}=\frac{1}{2}$

Which means that De l'Hopital rule works for this function, but $g'(x)$ changes its sign infinitely many times.

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    @AndréNicolas: Oh, I got it. I thought that it doesn't make sense because I could reduce numerator and denominator, but I see it's wrong. But I still can edit the post to make it useful :)2011-11-06