Talk:L'Hôpital's rule

Last limit

The last limit should be 1/2. Loisel 08:16 Feb 22, 2003 (UTC)

Yep, fixed it. Nice catch. Minesweeper 08:37 Feb 22, 2003 (UTC)

Typo in proof?

In the case when $|g(x)|\to +\infty$, shouldn't the formula read

${f(x) \over g(x)}={f(y) \over g(x)}+\left[1-{g(y) \over g(x)}\right]{f'(\xi ) \over g'(\xi )}$

instead of

${f(x) \over g(x)}={g(y) \over g(x)}+\left[1-{g(y) \over g(x)}\right]{f'(\xi ) \over g'(\xi )}$

(the numerator of the first term of the right hand side of the equation should be f(y) instead of g(y) imho) Kind regards, Pieter Penninckx

Yes it should. Even so, somebody should finish the second proof, there's a lot more that needs to be said.

Section 1 of the proof asserts:

According to Cauchy's mean value theorem there is a constant xi in c < xi < c + h such that:

   f'(xi) / g'(xi) =  ( f(c + h) - f(c) ) / ( g(c + h) - g(c) )

But the logic of this assertion does not seem correct to me. Cauchy's mean value theorem states:

  there is a constant Xi1 in  c < Xi1 < c + h
  such that f'(xi1) = ( f(c+h) - f(c) ) / h

  there is a constant Xi2 in  c < Xi2 < c + h
  such that g'(xi2) = ( g(c+h) - g(c) ) / h

So certainly,

     f'(xi1) / g'(xi2) = ( f(c+h) - f(c) ) / ( g(c+h) - g(c) )

However, you cannot assume that xi1 = xi2 !

"Hence Cauchy's mean value theorem ...it states that xi1 = xi2! And it's not proved like that." ~P. Y. from NTHU

I'm not saying the assertion is wrong, but I think the proof needs improvement.

• • Indeed, Cauchy's mean theorem CAN'T be derived that way. Instead it uses Rolle's theorem, see (http://en.wikipedia.org/wiki/Rolle%27s_theorem) and (http://en.wikipedia.org/wiki/Mean_value_theorem).

• • The proof of "With the indeterminate form infinity over infinity" is simply wrong. The correct proof can be found here

(http://planetmath.org/?op=getobj&from=objects&id=7611). The main text needs to be corrected.

Too strong requirement in overview?

Hello,

We were touched that $\lim _{x\to c}{f'(x) \over g'(x)}=A,A\in \mathbb {R} ^{*}$ requirement holds only for open interval (a,b) containing c (or with $b=\infty$ or $a=-\infty$)

Incorrect application of L'Hopital's rule in example image

The example below the infobox shows $\lim _{x\rightarrow {0}}{\frac {\sin {x}}{-0.5x}}=-{\frac {1}{2}}\lim _{x\rightarrow {0}}{\frac {\sin {x}}{x}}$. However, L'Hopital's rule cannot be applied to this limit, as the proof of the derivative of $\sin {x}$ depends on knowing precisely this limit.

Proof: ${\frac {d}{dx}}\sin {x}=\lim _{h\rightarrow {0}}{\frac {\sin {(x+h)}-\sin {x}}{h}}=\lim _{h\rightarrow {0}}{\frac {\sin {x}\cos {h}+\cos {x}\sin {h}-\sin {x}}{h}}=\cos {x}\lim _{h\rightarrow {0}}{\frac {\sin {h}}{h}}+\sin {x}\lim _{h\rightarrow {0}}{\frac {\cos {x}-1}{h}},\lim _{h\rightarrow {0}}{\frac {\sin {h}}{h}}=1,\lim _{h\rightarrow {0}}{\frac {\cos {x}-1}{h}}=0\therefore {\frac {d}{dx}}\sin {x}=\cos {x}$

I think this should be changed to a different example.

Singularities421 (talk) 21:54, 21 March 2021 (UTC)

L'Hopital's rule is not used to prove sin(x) ~ x. Valery Zapolodov (talk) 12:14, 10 November 2022 (UTC)

Then why is the visual example doing so? 128.138.65.161 (talk) 16:36, 6 August 2024 (UTC)

You mean $\lim _{x\to 0}{\tfrac {\sin x}{-0.5x}}=-2\lim _{x\to 0}{\tfrac {\sin x}{x}}.$ Also, you wrote ${\tfrac {\cos x-1}{h}}$ when you meant ${\tfrac {\cos h-1}{h}}.$ Also, I'd have preferred you didn't cram your entire proof into one line and make it uncomfortable to read.

Anyway, the equality ${\tfrac {\mathrm {d} }{\mathrm {d} x}}\sin x=\cos x$ can be proven without the use of the equality $\lim _{x\to 0}{\tfrac {\sin x}{x}}=1.$ This video[1] demonstrates one possible way to do so. Of course, it's rather loosey-goosey with the mathematics, but the basic idea is there to be developed into a rigorous proof.

On top of that, how do your statements imply that the application is incorrect anyway? It's not like we're trying to build mathematics from the ground up and have accidentally employed circular reasoning in doing so. All we're doing is solving a practical problem, so it's okay to take ${\tfrac {\mathrm {d} }{\mathrm {d} x}}\sin x=\cos x$ as a given. Like, what if we forgot about the limit thing, but we still remember the derivative thing? That's totally fine. ISaveNewspapers (talk) 16:08, 30 October 2024 (UTC)

Sorry about that first paragraph. It was quite unproductive and also a bit rude. ISaveNewspapers (talk) 16:14, 30 October 2024 (UTC)

Is the first L capitalized or not?

The article seems to flip-flop between L'Hôpital and l'Hôpital. The former is more common in this article, but when you check the article of the person the rule is named after (Guillaume de l'Hôpital), it uses the latter instead. Which way is it? ISaveNewspapers (talk) 16:18, 30 October 2024 (UTC)

The person himself generally spelled it "M. Le Marquis De l'Hospital" (and his given names were Guillaume-François-Antoine); though now typical, calling him "L'Hôpital" is kind of like calling the king of Spain by the name "Of Spain" or similar. In English, when used by itself people usually write "L'Hospital" or "L'Hôpital" (or historically "Lhospital" was also common), but when used along with a title or given name, the le is usually lowercased. –jacobolus (t) 17:05, 30 October 2024 (UTC)

Counterexamples: derivative of the denominator is zero

${\begin{aligned}{\frac {f'(x)}{g'(x)}}&={\frac {2\cos ^{2}x}{(2\cos ^{2}x)e^{\sin x}+(x+\sin x\cos x)e^{\sin x}\cos x}}\\&={\frac {2\cos x}{2\cos x+x+\sin x\cos x}}e^{-\sin x},\end{aligned}}$

which tends to 0 as $x\to \infty$, although it is undefined at infinitely many points.

I'm struggling to see how the latter fraction can be undefined at infinitely many points, this can happen only when denominator $2\cos x+x+\sin x\cos x$ is zero, which basically never happens as x approaches infinity. 46.13.221.87 (talk) 13:01, 12 November 2024 (UTC)

@46.13.221.87 Ah, the first unreduced fraction is undefined in infinitely many points, my bad 46.13.221.87 (talk) 13:05, 12 November 2024 (UTC)