# Does L'Hôpital's work the other way?

Hello fellows,

As referred in Wikipedia (see the specified criteria there), L'Hôpital's rule says,

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

As

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

Just out of curiosity, can you integrate instead of taking a derivative? Does

$$\lim_{x\to c}\frac{f(x)}{g(x)}= \lim_{x\to c}\frac{\int f(x)\ dx}{\int g(x)\ dx}$$

work? (given the specifications in Wikipedia only the other way around: the function must be integrable by some method, etc.) When? Would it have any practical use? I hope this doesn't sound stupid, it just occurred to me, and I can't find the answer myself.

## Edit

Take 2 functions $f$ and $g$. When is

$$\lim_{x\to c}\frac{f(x)}{g(x)}= \lim_{x\to c}\frac{\int_x^c f(a)\ da}{\int_x^c g(a)\ da}$$

true?

Not saying that it always works, however, it sometimes may help. Sometimes one can apply l'Hôpital's even when an indefinite form isn't reached. Maybe this only works on exceptional cases.

Most functions are simplified by taking their derivative, but it may happen by integration as well (say $\int \frac1{x^2}\ dx=-\frac1x+C$, that is simpler). In a few of those cases, integrating functions of both nominator and denominator may simplify.

What do those (hypothetical) functions have to make it work? And even in those cases, is is ever useful? How? Why/why not?

• You cannot use L'Hospital the other way to evaluate a limit. You can try an example, as I did and it just won't work. Realizing that taking a derivative of num and denom really means that you are linearizing and so that the ratio of the slopes (i.e. derivatives) will give you the answer to the limit, makes sense. But when you are anti deriving, what does that geometrically mean to the limit? I am not aware that "anti-L'Hospitaling" does anything... – imranfat Nov 22 '13 at 21:14
• @JMCF125: This is a good question. Perhaps you should clarify it by adding bounds to the integrals. Specifically, how about writing $\int_{c}^{x}\; f(t)\; dt$ instead of $\int f(x) dx$? Same for $g(x)$. I point out that neither answer addressed this question. – Chris K Nov 22 '13 at 21:18
• @ChrisK, good, then the constants are eliminated! Forgot about them... I'll add that ASAP- – JMCF125 Nov 22 '13 at 21:33
• @JMCP125 : based on Umberto P.'s answer, the answer seems to be "yes". The next question is, is this ever useful, like the "regular" L'Hopital's Rule itself is? – Stefan Smith Nov 23 '13 at 2:28
• @miracle173, I never said l'Hôpital's implied this. And I didn't say the integrals had to be real anti-derivatives; in fact, in the edit, I added ChrisK's suggestion, which clears that up. – JMCF125 Nov 24 '13 at 8:52

I recently came across a situation where it was useful to go through exactly this process, so (although I'm certainly late to the party) here's an application of L'Hôpital's rule in reverse:

We have a list of distinct real numbers $\{x_0,\dots, x_n\}$. We define the $(n+1)$th nodal polynomial as $$\omega_{n+1}(x) = (x-x_0)(x-x_1)\cdots(x-x_n)$$ Similarly, the $n$th nodal polynomial is $$\omega_n(x) = (x-x_0)\cdots (x-x_{n-1})$$ Now, suppose we wanted to calculate $\omega_{n+1}'(x_i)/\omega_{n}'(x_i)$ when $0 \leq i \leq n-1$. Now, we could calculate $\omega_{n}'(x_i)$ and $\omega_{n+1}'(x_i)$ explicitly and go through some tedious algebra, or we could note that because these derivatives are non-zero, we have $$\frac{\omega_{n+1}'(x_i)}{\omega_{n}'(x_i)} = \lim_{x\to x_i} \frac{\omega_{n+1}'(x)}{\omega_{n}'(x)} = \lim_{x\to x_i} \frac{\omega_{n+1}(x)}{\omega_{n}(x)} = \lim_{x\to x_i} (x-x_{n+1}) = x_i-x_{n}$$ It is important that both $\omega_{n+1}$ and $\omega_n$ are zero at $x_i$, so that in applying L'Hôpital's rule, we intentionally produce an indeterminate form. It should be clear though that doing so allowed us to cancel factors and thus (perhaps surprisingly) saved us some work in the end.

So would this method have practical use? It certainly did for me!

PS: If anyone is wondering, this was a handy step in proving a recursive formula involving Newton's divided differences.

• Very interesting, +1! I certainly didn't expect it to have been successfully used before. But why didn't you make the question you are answering when you found it worked? – JMCF125 Nov 27 '13 at 11:01
• Ah, after having reread your question, I think the technique you're asking about is a little different. In my case, I went from $f'/g'$ (where both $f'$ and $g'$ were non-zero) to $f/g$ (where both $f$ and $g$ were zero). I didn't use the reverse L'Hôpital to resolve an indeterminate form; I used it to make one. – Omnomnomnom Nov 27 '13 at 15:38
• However, when you apply the function $\omega_n(x)$ and cancel out $\Pi_{i=0}^n (x-x_i)$ (sorry, couldn't resist but to use big product notation :)), the "indeterminacy" disappears, it only appears as an intermediate step. Call $\omega'$ as $\alpha$ and $$\lim_{x\to x_i} \frac{\alpha_{n+1}(x)}{\alpha_n(x)} = \lim_{x\to x_i} \left(\lim_{c\to x}\frac{\int\alpha_{n+1}(z)\ dz}{\lim_{c\to x}\int\alpha_{n}(z)\ dz}\right)=x_i-x_{n}$$. Right? (didn't check it thoroughly, there's no comment preview) – JMCF125 Nov 27 '13 at 21:07
• Exactly! So, you could say that the trick in using L'Hopital's rule backwards is choosing the antiderivatives of $\alpha_{n+1}$ and $\alpha_n$ that give you the indeterminate fraction. I'm not sure what the ruole of $c$ is in your statement, though – Omnomnomnom Nov 27 '13 at 21:16
• About the $c$, it's to eliminate the constants, see the problem Arkamis referred and the answer of Kaz, which has a related idea. The point is to make the integral so small it would be alike a derivative, only as a different expression. (BTW, the second $c\to x$ limit was not suppose to be there, I pasted it twice) – JMCF125 Nov 27 '13 at 21:55

With L'Hospital's rule your limit must be of the form $\dfrac 00$, so your antiderivatives must take the value $0$ at $c$. In this case you have $$\lim_{x \to c} \frac{ \int_c^x f(t) \, dt}{\int_c^x g(t) \, dt} = \lim_{x \to c} \frac{f(x)}{g(x)}$$ provided $g$ satisfies the usual hypothesis that $g(x) \not= 0$ in a deleted neighborhood of $c$.

• Good, interpreted the question in the "right" way. – André Nicolas Nov 22 '13 at 21:26
• Or $\frac{\infty}{\infty}$. – marty cohen Nov 23 '13 at 2:56
• Very important, this only works if you know that $\lim \frac{f(x)}{g(x)}$ exists. Even if the first limit exists, you cannot conclude that the second one does. – N. S. Nov 23 '13 at 5:58
• @N.S., neither can you conclude the division of the derivatives exists, even when the division of the original functions does, as in the original l'Hôpital's. – JMCF125 Nov 23 '13 at 13:48
• @JMCF125 True, but the goal in L'H is: starting with a limit, you replace by a different limit. In L'H you replace the functions by the Derivatives, not the derivatives by the functions. In classical L'H, you check $\lim\frac{f'}{g'}$ and if this limit exists you are done. Here you check $\lim \frac{\int f}{\int g}$ and if the limit exists, you cannot say anything. Note that if you go the other direction, you just apply classical L'H – N. S. Nov 23 '13 at 15:40

No it does not, for example consider the limit $\lim_{x\to 0}\frac{\sin x}{x}$, L'Hôpital's gives you 1, but reverse L'Hôpital's goes to $-\infty$ (without introducing extra constants).

• +1 for being a simple, easy counterexample, although I assume that you're taking the integral of sin(x) to be -cos(x) - but that implies either a constant or a definite interval on the integral. – fluffy Nov 23 '13 at 4:33
• @fluffy : Yes, exactly, thank you – Arjang Nov 23 '13 at 4:43
• The conditions under which it works may be more restricted... – JMCF125 Nov 23 '13 at 17:20
• If we want to use reverse L'Hopital, it is our 'responsibility' to ensure that the newly integrated function is also of an indeterminate form, and that can be tackled easily using the integration constant. Here, instead of -cosx, one should use 1-cosx (in the numerator) for the limit to exist, and hence evaluate it to be 1. – arya_stark May 8 '18 at 8:08

Recall that the integral of a function requires the inclusion of an arbitrary constant. So, if $f(x) = x$, then $\int f(x)\ dx = \frac12 x^2 + C$.

Then, assuming your rule holds, then $$\lim_{x\to 0} \frac{x}{x} = \lim_{x \to 0} \frac{\int x\ dx}{\int x\ dx} = \lim_{x \to 0} \frac{\frac12 x^2 + C_f}{\frac12 x^2 + C_g} = \frac{C_f}{C_g}.$$

This means you can get literally any value you wish.

• What if we forget the constants? Wait, don't answer yet, I'll edit my question first. – JMCF125 Nov 22 '13 at 21:32
• @JMCF125 Note that it really makes no sense to talk of "forgetting the constants". For example, the following three functions are antiderivatives of $2\sin(x)\cos(x)$: $\sin^2(x)$, $-\cos^2(x)$, $-\cos(2x)/2$. Which one is the "correct" one to use if we are to "forget constants"? – Andrés E. Caicedo Nov 22 '13 at 21:59
• @AndresCaicedo, what if we use a definite integral $\displaystyle\int_0^x x\ dx=\frac{x^2}2$, as referred in my edit. That way, wouldn't the result be the same? – JMCF125 Nov 24 '13 at 8:57
• Why use $0$ as the lower bound. Also, it should be $\int_0^x t\ dt$, as it makes no sense to have the variable of integration be the same as a bound. In any case, a lower bound of $c$ makes some sense if $x \to c$, since, as others have said, one of the ways this question makes sense is if $f, g \to 0$ at $x=c$. But it elides the case of $f,g \to \infty$. – Emily Nov 24 '13 at 16:13
• @Arkamis, «(...) as it makes no sense to have the variable of integration be the same as a bound.», sorry, distraction. :S I meant for the top limit to be an $a$. I used the lower limit as $0$ so that in your example, the integral would be $\int^a_0 x\ dx=\frac{a^2}2+C-\frac{0^2}2-C=\frac{a^2}2$. Isn't it right like this? – JMCF125 Nov 27 '13 at 10:57

If the limit of the ratio of those functions exists, I suspect that it may be possible for a certain definite integral to have the same limit. Something like:

$$\lim_{x\to \infty}\frac{\int_x^{x+1}f(x)\ dx}{\int_x^{x+1}g(x)\ dx}$$

The idea is that we grow $x$, and compare the ratios of some definite integrals in the neighborhood of x, of equal width and position.

For instance, if we imagine the special case that the functions separately converge to constant values like 4 and 3, then if we take $x$ far enough, then we are basically just dividing 4 by 3. A definite integral of width 1 of a function that converges to 4 has a value that is approximately four, near a domain value that is large enough.

But this is like a derivative in disguise. If we divide a definite integral by the interval length, and then shrink the interval, we are basically doing derivation to obtain the original function that was integrated. Except we aren't shrinking the interval, since we don't have to: the fixed width 1 becomes smaller and smaller in relation to the value of $x$, so it is a "quasi infinitesimal", so to speak.

The real problem with this, even if it works, is that integrals don't seem to offer any advantage. Firstly, even if the function has the properties of being integrable, it may not be symbolically integrable, or it may be hard to integrate. Integration produces something more complex than the integrand.

The advantage of L'Hôpital's Rule is that we can reduce the power of the functions. If they are polynomials, they lose a degree in the differentiation, which can be helpful, and gives us a basis for instantly gauging the limit of the ratios of polynomials of equal degree simply by looking at the ratios of their highest degree coefficients. And then certain common functions at least don't grow any additional hair under differentiation, like sine, cosine, e to the x.

• Interesting, +1. But what about those exceptional functions whose derivative is more complicated? Wouldn't the integral ease calculation? – JMCF125 Nov 23 '13 at 13:55

L'Hospital's Rule is really a compact way of looking at ratio of the Taylor's polynomial for the numerator and denominator functions (w/o remainder). This polynomial is an exact fit to the function and its derivatives at the indicated point and close in a small neighborhood thereof.

When the limit is 0/0 that means that both Taylor's polynomials have 0 constant terms. So then you look at the x terms (which represent the derivatives). If you can pick out the limit from there, good -- if they are both zero you go on to the $x^2$ terms (if any).

So L'Hospital's Rule isn't some vague thing that was plucked out of the air and happens to use derivatives. Any method of evaluating limits ought to be based on a clear mathematical derivation. You can use integrals or gamma functions or field theory or anything that pleases you, as long as you show you have a sound basis for evaluating the limit that way.

• Can you help me finding such a basis? – JMCF125 Nov 28 '13 at 12:26
• @ Betty Mock What is exact permissible basis to modify the vanishing numerator and denominator ? – Narasimham Nov 20 '14 at 20:11
• Expand the top and bottom in a Taylor's polynomial. If the first term of both polynomials is zero, then you look at the second term. If both of those are zero, you go on to the third term, etc. Sooner or later you get a term which is not zero, and that is the one which shows you the limit. – Betty Mock Nov 22 '14 at 2:46

$$\lim_{ x \to 0} \frac{ sin \,x \,}{\,x}= \lim_{ x \to 0} \frac{\int sin \,x \,\ dx}{\int x\ dx} = \lim_{x \to 0} \frac{\ (- cos \, x)}{( x^2/2)-1} =\lim_{x \to 0} \frac{\ (-cos \, x)+1}{( x^2/2)} = 1$$

Constants should be appropriately chosen from function/ power series expansion of the integrands and distributed between numerator and denominator maintaining the balance.

The statement is vague but I am sure it can be properly worded/stated to include even second, third integrands in the numerator and denominator.

More simply:

When $\frac yx =const = m$ as L'Hospital limit, it is =$\frac{dy}{dx}$ by Quotient Rule.

$m = \frac {dy}{ dx} = \frac {y \, \pm C_1 }{ x} =\frac {y \, }{ x \pm C_2}.$