Converse of L'Hospital

Question

L'Hospital's Rule: If ${f(x)}\over{g(x)}$ is either $0\over0$ or $\infty\over\infty$, then the $\lim_{x\to \infty}$${f(x)}\over{g(x)}$ $=$ $\lim_{x\to \infty}$${f'(x)}\over {g'(x)}$. What I want to know is if the converse is true.

In other words, if $\lim_{x\to \infty}$${f(x)}\over{g(x)}$ $=$ $\lim_{x\to \infty}$${f'(x)}\over {g'(x)}$, then is ${f(x)}\over{g(x)}$equal to one of $0\over0$ or $\infty\over\infty$? By this I mean that there does not exist a case in which L'Hopital does not apply yet gives the correct answer.
It is not a repeat of another question, as many have suggested. I would ask those people to take a look at the question that they have said I have repeated. These are two different questions, and it surprises me that they could have been confused.

there are 7 indeterminate forms out of which L.H rule can be applied to only two ?(guess what) — , Mar 16 '18 at 17:03
I think this post is really worth reading and pretty much sums it all up. Answer by B. Cipra: https://math.stackexchange.com/questions/1710786/why-does-lhopitals-rule-fail-in-this-case/1710798#1710798 — imranfat, Mar 16 '18 at 17:15
Read the final sentence of this question. It's not a duplicate. The OP isn't asking about situations where L'Hopital might seem to apply but doesn't; he's asking if there can be situations where L'Hopital manifestly does not apply but happens to give the right answer anyway. — Micah, Mar 18 '18 at 00:43
@DaveL.Renfro hello Dave, I know it's very old post... But can you address me again to the post you mention in your comment. (The link seem not to point to it.) Thanks! — dfnu, Feb 01 '24 at 20:57
@dfnu: Searching in google's archive of the Usenet group "sci.math" -- I selected the top hit for the google search sci.math + google, then searched within there for Renfro + L'Hopital, then looked for the closest matching date -- led me to what is almost certainly the 10 January 2006 sci.math post I previously cited. The post is overkill for the OP's question and it was (continued) — Dave L. Renfro, Feb 02 '24 at 10:44
only intended to give a reference for more nuanced and advanced issues related to the OP's question, for those who might be interested. About a year ago I posted some more references in mathoverflow that I've known about since the early 1990s but don't believe I've ever posted in an online math group before -- see my comments to this answer. — Dave L. Renfro, Feb 02 '24 at 10:44
@DaveL.Renfro yes I had reached it already and found it very interesting. I'm reading right now Young's paper! Do you know of any work on when the converse of de l'Hospital Theorem holds? Thanks again! — dfnu, Feb 02 '24 at 13:41
@dfnu: Unless I'm misunderstanding what you mean by the/a converse of L'Hopital's theorem, the most natural converse is an immediate consequence of the generalized liminf-limsup version $\liminf \frac{f'}{g'}\leq\liminf\frac{f}{g}\leq\limsup\frac{f}{g}\leq\limsup\frac{f'}{g'},$ since if $\frac{f'}{g'}$ exists (finitely or infinitely), then we have $\liminf \frac{f'}{g'}=\limsup\frac{f'}{g'},$ from which it follows that $\liminf\frac{f}{g}=\limsup\frac{f}{g},$ and hence $\lim \frac{f}{g}$ exists (respectively, finitely or infinitely). — Dave L. Renfro, Feb 03 '24 at 19:21
@DaveL.Renfro Sorry I did not formulate my question properly. What I mean is: are there any conditions on $f$ and $g$ under which we can state that if the limit $f'/g'$ does not exist, then neither does the limit of $f/g$? — dfnu, Feb 03 '24 at 20:19
@dfnu: I don't know of any off-hand. I'll try to look further when I have the time. — Dave L. Renfro, Feb 03 '24 at 21:19

score 1 · Answer 1 · edited Feb 08 '20 at 06:40

1

Let $f(x)=x+\sin(x), g(x)=x$.

Then $$\lim_{x \to \infty} \frac{f(x)}{g(x)} =2$$ by the squeeze theorem but $$\lim_{x \to \infty} \frac{f'(x)}{g'(x)} =\lim_{x \to \infty} 1+\cos(x) = DNE$$

edited Feb 08 '20 at 06:40

Masacroso

30,417

answered Mar 16 '18 at 17:11

N. S.

132,525

score 1 · Accepted Answer · answered Mar 17 '18 at 15:24

Notice that $$\lim_{x \to \infty} \frac{1}{x^2} = \lim_{x \to \infty} \frac{0}{2x} = 0$$ even though the initial limit did not have an indeterminate form. That is, it is possible that you could get the right answer by applying L'Hopital even in cases where it is invalid. You could think of this as the calculus equivalent of the fake proof that $$ \frac{16}{64}=\frac{1}{4} $$ where you cancel out the sixes.

Converse of L'Hospital

2 Answers2