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L'Hôpital's rule for the indeterminate form $\frac00$ at finite points can be given a nice intuitive explanation in terms of local linear approximations. See for instance this textbook or this one. And for limits as x approaches $\pm\infty$, we can just let $u = 1/x$ and take $\lim_{u\to 0^+} \frac{f(1/u)}{g(1/u)}$, use the ordinary L'Hôpital's rule to get $\lim_{u\to 0^+} \frac{f'(1/u) \cdot(-1/u^2)}{g'(1/u)\cdot(-1/u^2)}$ and cancel the $(-1/u^2)$'s.

However, I've looked through half a dozen textbooks and not found even the barest hint of an explanation for why L'Hôpital's rule should also apply to the indeterminate form $\frac\infty\infty$. Even textbooks (like those linked above) that give an intuitive explanation for the $\frac00$ case simply state the $\frac\infty\infty$ case as an unjustified fact. Is there any intuitive explanation or (perhaps nonrigorous) calculation that explains this?

The best I've been able to come up with is

$$ \lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{1/g(x)}{1/f(x)} = \lim_{x\to a} \frac{-g'(x)/g(x)^2}{-f'(x)/f(x)^2} = \lim_{x\to a} \frac{f(x)^2 g'(x)}{g(x)^2 f'(x)} = \left(\lim_{x\to a} \frac{f(x)}{g(x)}\right)^2 \left(\lim_{x\to a} \frac{g'(x)}{f'(x)}\right)$$

using L'Hôpital for $\frac00$ and assuming that the limits of $f/g$ and $g'/f'$ exist. Then if both these limits are also nonzero, we can cross-cancel to get $\lim (f/g) = \lim(f'/g')$. I'd prefer a more geometric explanation, but the real problem is that the assumptions that $\lim(f/g)\neq 0$ and $\lim(g'/f')$ exists fail even in a simple case like $\displaystyle\lim_{x\to \infty} \frac{x}{e^x}$.

A more rigorous form of this argument was discussed on MSE, with the conclusion that it's not possible to deduce the $\frac\infty\infty$ rule from the $\frac00$ one in general. I'm not looking for a rigorous proof, so I'm hoping the situation may be easier. Can anyone give an informal justification of the $\frac\infty\infty$ rule that applies when the limit equals 0 and would be comprehensible to calculus students?

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    $\begingroup$ I use an example like $f(x)/g(x)=e^x/x$. Students intuitively get that $e^x$ and $x$ both grow to $+\infty$, but the exponential function grows "much faster". "How much faster?", I ask. "Let's compare their growth rates." This is what the ratio $f'(x)/g'(x)$ shows: it confirms our intuition that the rate at which $e^x$ is growing to infinity is much, much faster than the rate at which $x$ goes to infinity. (Flipping this example upside down gives you one where the limit is 0, as you want.) $\endgroup$ – Brendan W. Sullivan Nov 25 '19 at 22:43
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    $\begingroup$ Not sure if this is what you're looking for, but I have an explanation in this book lightandmatter.com/fund , secs. 6.3.2 and 6.3.4. $\endgroup$ – Ben Crowell Nov 26 '19 at 3:37
  • $\begingroup$ @MikeShulman: I wasn't sure quite what you were looking for, so let me know if my comment is on the right track and I'll expand it into an answer. $\endgroup$ – Brendan W. Sullivan Nov 26 '19 at 14:59
  • $\begingroup$ @BrendanW.Sullivan Thanks, but I'm hoping for a justification, not just an appeal to intuition. In particular I'd like it to include an explanation of why the exponential function grows much faster than a polynomial. $\endgroup$ – Mike Shulman Nov 27 '19 at 0:02
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    $\begingroup$ @BenCrowell Your explanation is the same as the argument I gave. Do you have some way around the inversion problem? $\endgroup$ – Mike Shulman Nov 27 '19 at 0:02
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Your approach is in fact geometrical, if we see all this happening on the Riemann sphere, i.e., the one-point compactification of ${\bf C}={\bf R}^2$. Briefly, we have the usual coordinate $z$ on ${\bf C}$, and the coordinate $u=\displaystyle{\frac{1}{z}}$ on a neighborhood of point at infinity. Calculus on this neighborhood is conducted in terms of the coordinate $z$ by the transformation $du=-\frac{1}{z^2}\, dz$.

Now in the case of $\displaystyle{\lim_{z\to a}f(z)}=\infty$ and $\displaystyle{\lim_{z\to a}g(z)}=\infty$, with certain assumptions on $f$ and $g$, we can think of $f(z)$ and $g(z)$ as coordinates around the point at infinity. In this framework, your derivation of the L'Hôpital rule for the case $\infty/\infty$ is quite geometrical.

The reason why a function like $h(z)=e^z$ does not fit into this framework is because $h$ has an essential singularity at infinity, so $h(z)$ fails spectacularly as a coordinate in a neighborhood of infinity.

Of course I understand that an instructor cannot take this geometrical framework into a calculus class and expect it to be helpful. But this does explain why your approach is the correct and natural approach.

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This is from a story: The Hodja was escorting Diba and Rushkady to a pancake-eating contest.

"Look", said Diba. "Let's say you get a head start at the pancake eating contest. If I eat pancakes twice as fast as you, eventually I'll have eaten almost twice as many pancakes as you'll have done."

"You'll never eat more pancakes than me!" Rushkady shouted.

"I meant hypothetically. If a function grows at twice the rate as another, it will approach being twice as great."

"Hmm. I guess I can see it when the functions are linear. If, say, the number of pancakes I eat is $f(t)=2t$ and the number you eat is $g(t) = t + 100$ (the $100$ being from your head start!), then the limit of $2t/(t+100)$ is $2$ as $t$ goes to infinity. But are you sure it works when the rate isn't constant?"

Diba thought a while. "Yes, I think so. If you eat pancakes faster at some times and slower at others, as long as I eat pancakes exactly twice as fast as you, I'll eventually approach eating twice as many. I mean, every time you eat one, I'll eat two. If you eat ten, I'll eat twenty. And so on. When you've $t$ pancakes beyond your head start, I will have eaten $2t$, so it's just like the first case."

"I guess. But you're saying that the limit of $f(x)/g(x)$ is the same as the limit of $f'(x)/g'(x)$. One might not be eating exactly twice as fast as another."

"OK, suppose $f'(x)/g'(x)$ is increasing toward $2$ as a limit. Then eventually $f'(x)/g'(x)$ will be at least $1.9$. That means $f(x)/g(x)$ will approach at least $1.9$. But eventually $f'(x)/g'(x)$ will be greater than $1.99$ and remain so. Then $f(x)/g(x)$ will approach at least $1.99$, in the long run. And so on, toward $2$."

"Oh. I see now. The limits will be the same."

"I don't know. Somehow I think it only works if the limit of the derivatives exists," said The Hodja.

But the bus had arrived at the pancake eating contest, and Diba's and Rushkady's thoughts turned to wondering whether either of them could eat half as fast as The Hodja.

The Hodja's comment is illustrated by $$\underset{t\to \infty }{\text{lim}} \frac{t+\cos t} {t+\sin t}\,.$$ I suppose your textbook has its own examples.

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We are looking for a rule to give us L'Hosptial's rule that says that $\frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)}$ when $f(x) = \infty$ and $g(x) = \infty$. So, let's simply say, "what if we were asked about the opposite, $\frac{g(x)}{f(x)}$?" Clearly, $\frac{g(x)}{f(x)} = \frac{1}{\frac{f(x)}{g(x)}}$. So, if we can find $\frac{g(x)}{f(x)}$ we can just flip it.

So, what is $\frac{g(x)}{f(x)}$?
$$\frac{g(x)}{f(x)} = \frac{\frac{1}{f(x)}}{\frac{1}{g(x)}} = \frac{\frac{1}{\infty}}{\frac{1}{\infty}} = \frac{0}{0}$$ That means that we can use L'Hospital to find $\frac{g(x)}{f(x)} = \frac{g'(x)}{f'(x)}$. Since $\frac{g(x)}{f(x)} = \frac{1}{\frac{f(x)}{g(x)}}$ then we can simply flip the fraction and get $\frac{f(x)}{g(x)} = \frac{f'(x)}{g'(x)}$.

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    $\begingroup$ L'Hopital's rule depends on a particular decomposition of a function as a quotient. So getting $0/0$ from $\frac{1/f}{1/g}$ only justifies applying L'Hopital to that quotient, as I did, and not to the different decomposition of the same function as a quotient $g/f$. $\endgroup$ – Mike Shulman Dec 13 '19 at 21:16
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We observe the proof on Wikipedia. The intuition behind it is Cauchy's mean value theorem, which states that

$$\frac{f(a)-f(b)}{g(a)-g(b)}=\frac{f'(c)}{g'(c)}$$

for some $c$ between $a$ and $b$. From here all we need to do is divide by $g(a)/g(a)$ to get

$$\frac{\frac{f(a)}{g(a)}-\frac{f(b)}{g(a)}}{1-\frac{g(b)}{g(a)}}=\frac{f'(c)}{g'(c)}$$

and let $a\to x$, where $\lim_{a\to x}|g(a)|=\infty$. This gives us

$$\lim_{a\to x}\frac{f(a)}{g(a)}=\frac{f'(c)}{g'(c)}$$

for some $c$ between $b$ and $x$. But by taking $b\to x$, we get the desired result by continuity/squeeze. In my opinion, I find this to be reasonably intuitive, while also being rigorous.

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