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If in L'Hôpital's rule we have that: $f,g :(a,b)\to \mathbb{R}$, there exist $f'(x)$, $g'(x)$, and $g'(x)\ne0$,

$$\lim_{x\to a^+} \frac{f(x)}{g(x)} = L,$$

and also $\lim_{x\to a^+} f(x)=\lim_{x\to a^+} g(x) = 0$, must it be also

$$\lim_{x\to a^+}\frac{f'(x)}{g'(x)} = L$$

or not? I think it is wrong in some cases, but I can't find an example.

Angelo
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Steel_Rat11
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  • $f(x)=\sin(x)$, $g(x)=x$, $a=\infty$. – Andrés E. Caicedo Dec 23 '13 at 18:28
  • I'm sorry, there is some additional conditions – Steel_Rat11 Dec 23 '13 at 18:32
  • I'm confused about what you're asking. – dfeuer Dec 23 '13 at 18:58
  • We know that L'Hôpital's rule is correct. Does correct that L'Hôpital's rule works at back way? – Steel_Rat11 Dec 23 '13 at 19:02
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    I think the question does not at all reflect what Steel_Rat has in mind. I believe he wants to know whether $\lim f'/g' = \lim f/g$ when $\lim f'/g'$ has indeterminate form (short answer: no.) – user7530 Dec 23 '13 at 19:59
  • Could someone tell me if the inverse L'Hôpital's rule works in the case that $f’(x)$ and $g’(x)$ are both monotonically increasing ? – Angelo Feb 06 '24 at 07:50
  • @Angelo convexity is not enough for deducing $f'/g'$ from $f/g$. I guess you already proved that this implication works if one of the two functions is a polynomial (per your comment to a more recent question). But you cannot further generalize. – dfnu Feb 10 '24 at 17:09
  • @dfnu, please, could you write a counterexample about the fact that convexity is not enough for deducing $f’/g’$ from $f/g;?;$ Actually I have already proved that this implication works if $,g(x)=x^p,$ where $,p\in\Bbb R^+$. – Angelo Feb 10 '24 at 18:42
  • @Angelo my example is really convoluted. I'm sure if you ask a question you'll get a much nicer answer from the rest of the community! – dfnu Feb 10 '24 at 20:52
  • @dfnu, please tell me about your counterexample, no matter how convoluted it may seem. – Angelo Feb 10 '24 at 21:23
  • @Angelo Too little time right now to type all the details. When you post a proper answer, if nobody gives you the example you require faster than me, I'll try to show you mine. In the meantime, I suggest you look into the details of your proof for monomials. You should see the main reason why the reverse of de l'Hopital works in this case and this may help you looking in the right direction for a more "pathological" situation. – dfnu Feb 12 '24 at 13:13
  • @Angelo did you make any progress in the meantime? – dfnu Mar 14 '24 at 20:44

1 Answers1

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The exception is that there may be a case that $\lim_{x\to a^+} f'(x)$ does not exist.

So let's define

$$g(x)=x, \quad f(x)=\int_0^x(1+\sin(1/y))dy$$.

Apparently $\lim_{x\to 0^+}g(x)=\lim_{x\to 0^+}f(x)=0$, and $\lim_{x\to 0^+} f'(x)$ does not exist.

The tricky part is to show that $\lim_{x\to 0^+} f(x)/g(x)=1$. But first of all, it is true intuitively because $\sin(1/y)$ oscillates infinitely fast so the integral cancels out.

More rigorously, we can treat the integral by define $z=1/y$ $$ \int_0^x\sin(1/y)dy=\int_{1/x}^\infty \frac{\sin(z)}{z^2}dz $$

It can be shown that $\int_\alpha^\infty[\sin(z)/z^2]dz$ is bounded by two sums $$ \sum_{n=N}^\infty a_n>\int_\alpha^\infty\frac{\sin(z)}{z^2}dz>\sum_{n=N}^\infty b_n $$

Where $N\equiv \lfloor \alpha/(2\pi)\rfloor$, $a_n$ and $b_n$ are defiend as $$ a_n\equiv \int_{2n\pi}^{(2n+2)\pi}\frac{\sin(z)}{z^2}dz, \quad b_n\equiv \int_{(2n-1)\pi}^{(2n+1)\pi}\frac{\sin(z)}{z^2}dz $$

$$ a_n=\int_{2n\pi}^{(2n+1)\pi}\sin(z)\left(\frac{1}{z^2}-\frac{1}{(z+\pi)^2}\right)dz $$ So $$ 0<a_n<\int_{2n\pi}^{(2n+1)\pi}\left(\frac{1}{z^2}-\frac{1}{(z+\pi)^2}\right)dz =\frac{2\pi^2}{2n(2n+1)(2n+2)\pi^3} $$ Thus $a_n\sim\mathcal{O}(1/n^3)$, and $\sum_{n=N}^\infty a_n \sim \mathcal{O}(1/N^2)=\mathcal{O}(1/\alpha^2)$

btw,the properties of the sum can be found on wolframalpha if you are interested. link 1 link 2

Similarly we can prove that $0>\sum_{n=N}^\infty b_n \sim \mathcal{O}(1/\alpha^2)$

So $$ \lim_{x\to0^+}\frac{1}{x}\int_0^x\sin(1/y)dy=\lim_{x\to0^+}\frac{1}{x}\int_{1/x}^\infty \frac{\sin(z)}{z^2}dz =\lim_{x\to0^+}\frac{1}{ x}\mathcal{O}(x^2)=0 $$

MoonKnight
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  • Nice answer! I like it very much. Could you tell me if the inverse L'Hôpital's rule works in the case that $f’(x)$ and $g’(x)$ are both monotonically increasing ? – Angelo Feb 06 '24 at 07:49