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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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    Yes, the denominator will converge to $0$, but it still doesn't mean that the limit cannot be determined. For example, denominator can be reduced with numerator - see my answer.2011-11-05
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    Hm, turns out I was attaching L'Hôpital's name to a restricted variant of the rule.2011-11-05
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    Now edited to work with the general formulation of L'Hôpital.2011-11-05
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    Thank you. So, we could conclude that the hypothesis regarding the sign of $g'(x)$ is "hidden" in the other hypothesis of the theorem? In other words, is it _necessary_ that $g'(x)$ cannot change its sign infinitely many times? If this is the case, we could conclude that the visual demonstration (which uses the fact that $g'$ must be definitely greater than 0) is "universal"?2011-11-05
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    I'm not _completely_ sure. I'm tacitly assuming that $g'$ is continuous in a neighborhood of $a$. It is possible for a function to be differentiable everywhere _without_ the derivative being everywhere continuous, but the standard examples all involve lots of zero crossings anyway.2011-11-05
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    Update: [Darboux' theorem](http://en.wikipedia.org/wiki/Darboux's_theorem_%28analysis%29) implies that we don't need to assume that $g'$ is continuous. So yes, you can assume that $g'$ is eventually positive.2011-11-05
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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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    There are several versions of L'Hospital's Rule. What most of them have in common is if $\lim \frac{f'(x)}{g'(x)}$ exists, then $\lim\frac{f(x)}{g(x)}$ exists (and is the same). In a sense, L'Hospital's Rule "works" in this example, because the limit of the ratios of the derivatives does not exist, since the ratio is not defined for $x$ arbitrarily close to $0$.2011-11-05
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    @AndréNicolas: Well, I used the version of l'Hopital rule which OP mentioned in his question. But why "the limit of the ratios of the derivatives does not exist" ? Their ratio is $\frac{1}{2} \forall x \not=0$2011-11-05
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    I see, this example works, but we don't need the De l'Hospital rule to calculate this limit. The limit argument is actually $1/2$, provided that $x\ne0$. I would like to find a "real" example, which needs the rule to be solved. (btw, there is an error in your derivative of $f(x)$)2011-11-05
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    @zar: This example is *just_to_show*. I am sure it can be changed to be so complicated,that you will "need" to use the rule, if you want so :) also, can't find the mistake2011-11-05
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    @GrigorGevorgyan the derivative of $-x\sin(1/x)$ is $-\sin(1/x)+\cos(1/x)/x$.2011-11-05
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    @zar: Oops, did't copy that part :) will edit now2011-11-05
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    @Grigor Gevorgyan: About the differentiation slip, which changes things, but not much: The derivative of $-x\sin(1/x)$ is $(1/x)\cos(1/x) -\sin(1/x)$. The derivative is still $0$ for some points in any deleted neighbourhood of $0$, since $\tan u=u$ has arbitrarily large solutions. So the ratio $f'(x)/g'(x)$ is undefined for some points in any deleted neighbourhood of $0$. Thus the limit of the ratio of the derivatives, by the usual definition of limit, does not exist.2011-11-05
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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