I understand perfectly well how to apply l'Hopital's rule, and how to prove it, but I've never grokked the theorem. Why is it that we should expect that $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x \to a}\frac{f'(x)}{g'(x)},$$ but only when specific criteria are met, like $f(x)$ and $g(x)$ approaching $0$ as $x \to a$? I've never found the proof of the theorem to shed much light on this (thought this might be because I don't have much of an intuitive sense as to why the Cauchy Mean Value Theorem holds, either). I've always been mystified as to how l'Hopital (Bernoulli?) ever discovered this theorem, beyond accidentally discovering that it holds in some cases and then generalizing it to a theorem. So, if you have any ideas as to where the theorem came from, and why it makes intuitive sense, I'd appreciate it if you shared them!
6 Answers
The intuition is that although both numerator and denominator tend to zero or infinity, what eventually matters is their respective rate of change. They do not approach zero or infinity at the same rate and thus the one with the highest rate of change dominates the other.
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4$\begingroup$ I like this short answer because this is what you intuitively do anyway in precalculus when you look at the limit of a fraction: See which part wins when changing (!) x. l'Hôpital's rule makes this rigorous. $\endgroup$– vonjdCommented Dec 8, 2016 at 11:54
There is a case where the result is obvious: Suppose that $f,g$ are continuously differentiable at $a$, that $g'(a)\ne0$, and $f(a)=g(a)=0$. Under these assumptions, $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{\lim_{x\to a}\frac{f(x)-f(a)}{x-a}}{\lim_{x\to a}\frac{g(x)-g(a)}{x-a}}=\frac{f'(a)}{g'(a)}. $$ Of course, under the assumption that $f',g'$ are continuous at $a$, the latter is also $\displaystyle\lim_{x\to a}\frac{f'(x)}{g'(x)}$. Note that this is just a particular case of the result, but it is one of the most used cases, and it is widely applicable, for example, it also holds of analytic functions on an open domain in the plane, a case definitely not covered by the usual (real valued) argument.
The proof of the theorem is an extrapolation of this case, actually: We note that it is not necessary to assume that $f',g'$ are continuous at $a$, or even that they are defined there, since the differences $f(x)-f(a)$, $g(x)-g(a)$ are related to derivatives via the mean value theorem, so if $f(a)=0=g(a)$, we have $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f(x)-f(a)}{g(x)-g(a)}=\lim_{x\to a}\frac{f'(t_x)}{g'(t_x)} $$ for some $t_x$ between $x$ and $a$, so if $f'(t)/g'(t)$ converges as $t\to a$, then so does the displayed fraction.
The key is really the mean value theorem. Once one sees that this is all is needed in the case $f(a)=g(a)=0$, the other cases can be foreseen, and the argument adapted to cover them as well.
As you say, l'Hôpital's rule is due to Bernoulli, see here. You may also be interested in these slides by Ádám Besenyei on the history of the mean value theorem. Together with the history of the result, the geometric intuition discussed there may help you find the result more intuitive as well.
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$\begingroup$ For the case of analytic functions, L'Hospital's rule is essentially trivial precisely because of the local power series expansions. What makes its truth in the real case subtle is that the functions need not have power series expansions at $a$. $\endgroup$– KCdCommented May 23, 2022 at 4:08
Around $x=a$ each of these functions can be approximated by their tangent line: $$ f(x)\approx f'(a)(x-a)+f(a) $$ $$ g(x)\approx g'(a)(x-a)+g(a). $$ Thus $$ \frac{f(x)}{g(x)}\approx \frac{f'(a)(x-a)+f(a)}{g'(a)(x-a)+g(a)}. $$ Taking the limit of the right hand side gives $\frac{f'(a)}{g'(a)}$.
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3$\begingroup$ This is great for the case where they both tend to 0. Does it work, or can it be extended, for the case where they both tend to infinity? $\endgroup$– ruakhCommented Nov 28, 2013 at 21:22
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1$\begingroup$ @ruakh use the transformation $y=\frac{1}{x} \rightarrow 0$ and expand $\frac{f(y)}{g(y)}$. $\endgroup$ Commented Nov 28, 2013 at 22:58
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$\begingroup$ Or, take $a(x)=f(x)^{-1}$ , $b=f(x)^{-1}$. You now want the limit $\frac{b(x)}{a(x)}$ $\endgroup$ Commented Nov 29, 2013 at 2:21
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$\begingroup$ @ja72: By "they both", I meant f and g. In other words -- this is great for indeterminacy of the 0/0 type, but I'm not sure if it works for indeterminacy of the ∞/∞ type. $\endgroup$– ruakhCommented Nov 29, 2013 at 22:54
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1$\begingroup$ @ruakh $\lim_{x \to a} \frac{f(x)}{g(x)}=\lim_{x \to a} \frac{\frac{1}{g(x)}}{\frac{1}{f(x)}}$ $\endgroup$ Commented Nov 29, 2013 at 23:18
Suppose $f(x),g(x)\to0$ as $x\to a$. Look at the point $(a,0)$ under a microscope. The graphs of $f$ and $g$ in that view look like straight lines. Recall that $dx$ is an infinitely small increment of $x$, that looks like $0$ unless you use this microscope. The slopes of those lines you see under the microscope are $$ \frac{f(a+dx)-f(a)}{dx} = \frac{f(a+dx)}{dx}\text{ and similarly }\frac{g(a+dx)}{dx}. $$ The ratio of their slopes is therefore $$ \frac{f(a+dx)}{g(a+dx)} $$ which is what in modern notation we call $$ \lim_{x\to a}\frac{f(x)}{g(x)}. $$ But it's the ratio of their slopes, so it's $\displaystyle\lim_{x\to a}\frac{f'(x)}{g'(x)}$.
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1$\begingroup$ I can confirm this answer. I reinvented the rule by myself on high school. I showed the theorem (for which I had no proof) to my math teacher and he was very suprised. I found the rule by picturing the graphs. It is clear that if you divide the two heights of the slopes, it will go to a finite number. Than the rule becomes clear. $\endgroup$– Lucas K.Commented Jan 13, 2014 at 22:23
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$\begingroup$ Can you please explain a little more what you mean by dividing the $\endgroup$ Commented Mar 3, 2020 at 18:44
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$\begingroup$ @RiyaVerma : You need to finish your sentence before I can understand it. $\endgroup$ Commented Mar 4, 2020 at 20:21
This is Joe's answer and Michael's answer but with pictures! Also this answer is for intuition not rigour.
We start with functions $y=f(x)$ and $y=g(x)$ that are differentiable and have roots at $x=a$. Something like (maybe $f(x)$ blue and $g(x)$ red)
Now as Michael says, when we zoom in on differentiable functions they look like lines --- near a point, functions are well-approximated by their tangents. How good is this approximation: doesn't matter here really!
So we draw the tangents:
Now $f(x)$ is light blue and $T_f$ is light blue. Also $g(x)$ is green and $T_g$ is red. At this point we could zoom in or we could follow Joe's argument. Instead I am going to translate everything to the origin!
In terms of your limit you can do this via $z=x-a$:
$$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{z\rightarrow 0}\frac{f(z)}{g(z)}:$$
Now zoom in (This should be a $z$-axis)... and of course $T_f(z)=f'(a)z$ and $T_g(z)=g'(a)z$ (where the derivatives are respect to $x$):
Now you have:
$$\frac{f(x)}{g(x)}=\frac{f(z)}{g(z)}\approx \frac{f'(a)z}{g'(a)z}=\frac{f'(a)}{g'(a)},$$
near $z=0\Leftrightarrow x=a$.
Maybe the translation is unnecessary if we write $T_f(x)=f'(a)(x-a)$ and $T_g(x)=g'(a)(x-a)$.
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2$\begingroup$ Very nice intuition; thank you for the graphs, they help a lot. $\endgroup$ Commented Mar 17, 2016 at 3:52
To simplify @Michael's answer even further, briefly put for infinitesimal $\epsilon$ we have that $f(\epsilon)$ is indistinguishable (in a suitable sense) from $\epsilon f'(\epsilon)$, and similarly for $g$, and therefore $\cfrac{f(\epsilon)}{g(\epsilon)}$ is indistinguishable from $\cfrac{\epsilon f'(\epsilon)}{\epsilon g'(\epsilon)}=\cfrac{f'(\epsilon)}{g'(\epsilon)}$. Since the OP was specifically interested in grokking this could perhaps help.
