Talk:L'Hôpital's rule

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Contents 1 Last limit 2 Location 3 Edit 4 Use examples that avoid the three naive rules 5 Other proofs 6 Error in example? 7 Typo in proof? 8 Too strong requirement in overview? 9 definition too ridiculous 10 More examples of other indeterminate forms would be helpful. 11 Story of the creator 12 Local Linearity Proof, Pending 12.1 Other proofs 12.2 Proof by local linearity 13 Name spelling 14 Problem in proof that L'Hôpital's can be applied to infinity over infinity 15 You Divided by zero 16 How to pronounce the name 17 Other applications example is poor example of L'Hopital's rule 18 Important error 19 explanation of revert 20 L'Hospital Proofs

[edit] 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)

[edit] Location

Stupid question time again... Should this be here, or at L'Hôpital's rule...? -- Oliver P. 02:30 Feb 26, 2003 (UTC)

I'm not sure if it's written somewhere, but the policy seems to be that the content should be placed under the title with no accents, umlauts, circumflexes, etc. and redirect titles with those characters to the main article. -- Minesweeper 06:33 Mar 9, 2003 (UTC)

Huh?? Since when?? What about Kurt Gödel and Paul Erdős? There are thousands of these and I've never heard anyone say a single word against them. Michael Hardy (talk) 02:37, 20 April 2008 (UTC)

[edit] Edit

I have to wonder: as I did my first time major edit, I expected my proof of l'hopital's rule to be torn apart and totally rewritten. I have not yet recieved a single edit after mine. What's up? The edit was good, the proof was correct? Or does nobody care about this corner of wikipedia? --Sverdrup 16:19 Nov 18, 2003 (UTC)

Actually, the history shows there have been edits after yours. I'm guessing L'Hôpital's rule probably isn't one of the most frequently visited pages in Wikipedia though, so it's likely an edit that isn't vandalism would go unnoticed for a while in something as specialised as this. My only thought was - is there a point to the strange purple box? Angela 19:47, 20 Nov 2003 (UTC)

I looked at another page displaying a proof, and that article had a purple box around it. It may not be convention, and it may not be beautiful, but I'll let someone else decide. Sverdrup 16:24, 21 Nov 2003 (UTC)

l'Hopital's rule should be cross-referenced with Stolz-Cesàro theorem, because they are similar in flavour. It is natural to think about a' for sequences as a'_n=a_{n+1}-a_n / 1.

I've added a link to that from this article. Michael Hardy 04:08, 2 April 2007 (UTC)

[edit] Use examples that avoid the three naive rules

Here are three naive rules that are erroneous and could plausibly be believed by anyone who might be struggling to understand this topic:

(1) The numerator approaches 0; therefore the quotient approaches 0.

(2) The denominator approaches 0; therefore the quotient approaches ∞ or −∞.

(3) The numerator and denominator both approach the same number (0); therefore the quotient approaches 1.

To avoid reinforcing errors, the earliest examples involving the indeterminate form 0/0 should therefore be cases in which the limit is neither 0 nor 1 nor ∞ nor −∞.

Does anyone have a good example of this involving the indeterminate form ∞/∞? Michael Hardy 00:01, 13 Sep 2004 (UTC)

I added an example of 0/0 where the limit is 69.5. By the way, I think the first two rules above are correct if you meant "Only the numerator" and "Only the denominator". Eric119 01:03, 13 Sep 2004 (UTC)

Unfortunately, the example you added was one that could be done so easily without L'Hopital's rule that it's not really a good example. And besides, using L'Hopital's rule on problems of that kind often runs the risk of the same kind of circular reasoning that is referred to in the article. (Of course, the proposed "naive rules" are correct with the qualifications you mention; that is the reason people may be tempted to follow them.) Michael Hardy 20:20, 13 Sep 2004 (UTC)

[edit] Other proofs

I don't understand the proof by the linearity argument (I don't even know what does it mean). Can someone please expand it (or explain it to me on this talkpage) so that it's understandable?

Maybe I'll add something there at some point... Michael Hardy 19:09, 20 Jan 2005 (UTC)

Additionally, the seccond sentence assumes f(x) tends to infinity if g(x) tends to infinity,

I don't see how you get that from the second sentence in that proof. Michael Hardy 19:09, 20 Jan 2005 (UTC)

which there is no reason to assume (or it must be somehow explained if there is some). --hhanke 21:57, 19 Jan 2005 (UTC)

Local linearity: a differentiable function is "locally linear," meaning that it is practically linear if a sufficiently small part of it is viewed.

The argument is that if tends to 0/0, then . Then, provided the functions are differentiable at 0, each function can be approximated locally: e.g., for f(x), it can be approximated near 0 by the line with slope f'(0) that passes through (0,0).

So we have and , both near x = 0.

Hence .

I hope that works for a quick explanation. -Rjyanco 19:17, 20 Jan 2005 (UTC)

[edit] Error in example?

I was looking at the example for lim {x->0+} (x ln x) which is the next to last example, and I was wondering if it has some errors in it:

first, I don't see how lim {x->0+} (x ln x) = lim (x->0+) ((ln x)/x)? I've tried doing the math and don't see how x ln x = (ln x)/x directly or by its derivatives

second, assmuming the first equation is true, then after applying l'hopital's we get lim {x->0+} ((1/x)/1) = 0. Well, lim {x->0+} ((1/x)/1) = lim {x->0+} (1/x) and for all intensive purposes i thought lim {x->0+} (1/x) = oo+ (positive infinity)? It seems to me someone applied l'hopital's another time and got lim 0/1 = 0 but you can't do that since we don't have a 0/0 or OO/OO condition. If you graph 1/x you clearly see it going to oo+ as x->0+. [stux]

I think you're right. I believe I have fixed it now. Eric119 22:55, 8 Feb 2005 (UTC)

(Since non-native-English-speaking people may be reading this, please not: "for all intents and purposes" is a standard locution; "for all intensive purposes", appearing above, is a jocular parody.) Michael Hardy 22:47, 9 Feb 2005 (UTC)

[edit] Typo in proof?

In the case when , shouldn't the formula read

instead of

(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.

[edit] Too strong requirement in overview?

Hello,

We were touched that requirement holds only for open interval (a,b) containing c (or with or )

[edit] definition too ridiculous

The definition of L'Hopitals rule is too rediculous. In other words its not accessible. The formal definition either needs to be toned way down, or come later. I will try to do the ladder for now. Fresheneesz 01:44, 5 June 2006 (UTC)

it seems as though your proof relies on lim f(x)=f(x) as x aproches a, but this would only be true if the theorm required continuity for [a,b] which it doesn't.

[edit] More examples of other indeterminate forms would be helpful.

I came here to see examples for the forms , , and 0⁰. Found none.. —The preceding unsigned comment was added by 84.108.154.59 (talk • contribs) .

Those are not forms that L'Hopital's Rule is explicitly set up to solve. You can usually solve them using natural logarithms to turn them into one of the other standard forms. Anyone think the article should have this sort of trick added to it? -- dcclark (talk) 18:15, 14 July 2006 (UTC)

• Not I .   --Bob K 18:35, 14 July 2006 (UTC)

[edit] Story of the creator

My univeristy professor told us that the origial author sold the rule to Hospital, because he was apparently poor and Hospital wanted to be remmembered. As I have no citation or source, I'll wonder if anyone else heard something like that.. Lovok 12:16, 15 August 2006 (UTC)

i did

MathWorld says that l'Hôpital published it in 1696, but acknowledged the Bernoulli brothers. A letter by John Bernoulli gives the rule and a proof, so it seems likely he discovered it. Tmartin 21:00, 28 November 2006 (UTC)

I think this is great. I consider it an important early example of "the purchaser overshadowing the creator". I happen to have a text by Professor Earl D. Rainville called Unified Calculus and Analytic Geometry from 1961. It's interesting that he presents this as "Theorem 36" on page 262, rather than L'Hopital's Rule. (He also presents it late, after the student has already seen significant material. More modern texts seem to hurry to put it out there within the first 30 pages of the presentation of Derivatives.)TaoPhoenix 06:57, 21 August 2007 (UTC)

[edit] Local Linearity Proof, Pending

There is a section in the article, which goes as follows:

[edit] Other proofs

There are more intuitive proofs of the rule. If

tends to the indeterminate form 0/0, then the rule can be proven with a local linearity argument. If it tends to the indeterminate form , then this can be converted to 0/0 form using the identity : By assuming this limit equals L, and taking the derivative of the numerator and denominator, it can be proven that .

• • • What follows is a local linearity proof that, in a sense, is something that I came up with while thinking about the problem. If it works and you (reader) have no objections to it, I'll include it soon. And I quote:

[edit] Proof by local linearity

Suppose that functions f(x) and g(x) are continuous and differentiable on the interval (a,b). In addition, there is a function h(x) that is equal to the ratio of f(x) and g(x). In other words,

When these functions are equal to a real number c, which is on the interval (a,b), their ratio

For example, if f(x) = x³ + 1 and , then h(x) evaluated at 1 is

However, suppose that , where c is again a value on the interval (a,b). Then

0 divided by 0 is indeterminate. This means that h(c) could be any real number or infinity, based on the context. However, f(x) and g(x) are continuous and differentiable at x = c, they can be approximate by lines (because of local linearity.

The linear approximation (or linearization) of f(x) and g(x) around x = c is, respectively, and . The function h(x) near x = c, then, can be approximated by

Using the mathematical expression for the functions' linearizations,

Since f(c) and g(c) and 0 regardless of the value of x,

There is factor of (x − c) in both the numerator and denominator, which can be canceled, revealing

If f'(c) and g'(c) are both real and both are not zero, then

This concludes the proof. If a 0 over 0 form occurs again, L'Hôpital's rule can be used again (unless f(x) and g(x) resemble the zero function y = 0 around c).

The over can be treated similarly, when . Since

which transforms a over form into a 0 over 0 form, which can be evaluated with L'Hôpital's rule.

• • • (end section) All right. That's the proof. If no one disagrees, and please disagree or suggest changes if you don't like it, I'll put it on the L'Hôpital's rule page soon. --Gracenotes ^T § 17:01, 28 October 2006 (UTC)

• • • • The key statement here is: The function h(x) near x = c, then, can be approximated by

This is equivalent to L'Hopital's rule, and it's not proved at all. Septentrionalis 23:20, 30 October 2006 (UTC)

[edit] Name spelling

"l'Hôpital's rule (alternatively, and quite incorrectly, l'Hospital's rule)"

Sources for this? My understanding is that the two are almost equivalent in French and, in any case, the latter is a transliteration. This is like saying it's incorrect to spell Paul Erdős as Paul Erdos. In 'Mathematical Methods for Science Students' l'Hospital is used, presumably for a lack of a circonflexe at the printer. I don't see anything objectionable to the spelling and I think it's certainly a bit excessive to say that it's used "quite incorrectly". It may not be how l'Hôpital would spell his own name, but it's an acceptable transliteration of which I'm sure he would approve Tmartin 21:00, 28 November 2006 (UTC)

The circumflex character in the word "hôpital" denotes an unpronounced letter s. If the s is not pronounced in French (hence the circumflex), why should it be added back in when transcribed into English, such that English speakers will pronounce it? Most native-English speakers do not know what a circumflex is, let alone that the s it replaces is not pronounced. There's no legitimate reason not to render it with the circumflex--Wikipedia is full of diacritical marks, and IPA pronunciation similarly abounds. His name should be spelled using the French character set (i.e., l'Hôpital). 216.145.255.2 11:31, 5 January 2007 (UTC)

Well, as the article on Guillaume de l'Hôpital himself says:

l'Hôpital is commonly spelled as both "l'Hospital" and "l'Hôpital." The Marquis spelled his name with an 's'; however, the French language has since dropped the 's' (it was silent anyway) and added a circumflex to the preceding vowel.

In other words, the modern French spelling is Hôpital, and the old French spelling was Hospital. The spelling may be older, but it was still current in l'Hôpitale's time. So really, while it seems perfectly fair to say that "Hôpital" should be the prefered spelling in Modern English, it is absurd to call Hospital incorrect. --Iustinus 00:41, 31 January 2007 (UTC)

I agree in fact both spellings are common in mathematical literature and from pure language point of view I think you can argue both versions too (common spelling vs unchanged spelling of names). So I took the liberty to remove the incorrectly phrase.

[edit] Problem in proof that L'Hôpital's can be applied to infinity over infinity

In the proof that L'Hôpital's theorem can be applied to infinity over infinity, the theorem is applied to

but nowhere was it assumed that is non-zero throughout as needed in order to apply the theorem for the case

Micha

It is required. It follows from the existence of the limit. 80.178.152.155 20:51, 18 April 2007 (UTC)

[edit] You Divided by zero

0/0? You can't divide by zero if you do it gives you all sorts of weird answers. —The preceding unsigned comment was added by 71.246.231.88 (talk) 23:04, 2 January 2007 (UTC).

Hence l'Hôpital's rule, my friend. -216.145.255.2 11:32, 5 January 2007 (UTC)

We're talking about limits here. We can go as close as we want to 0 without actually touching it.130.234.198.85 23:43, 18 September 2007 (UTC)

[edit] How to pronounce the name

I guess most of our readers are not fluent french speakers, so it would be cool to have instructions in the article. Both IPA version and an english approximate would improve the article.130.234.198.85 23:57, 18 September 2007 (UTC)

[edit] Other applications example is poor example of L'Hopital's rule

The example given under "other applications" as an example of an indeterminate inf/inf example is a poor example of L'Hopital's rule for 2 reasons: 1. It is easier solved without using L'Hopital's rule 2. You can take the directive of denominator repeatedly and it will remain an inf/inf indeterminate form. It doesn't terminate. This is represented by the term (2x-1)/(2*sqrt(x^2-x)) whose lim as x-> inf is declared to be 1. [How? By factoring an x out of the numerator and denominator to yield (2x(1-1/x))/(2x(sqrt(1-1/x))), canceling the 2x yields lim x-> inf (1-1/x)/sqrt(1-1/x) which is 1.] Refering back to item (1), if you apply this very same step ('pulling' an x^2 out of the sqrt and canceling) before applying L'Hopital's rule you get the same answer. L'Hopital's rule isn't needed AND doesn't get you any closer to the solution. It just complicates the problem. —Preceding unsigned comment added by 71.212.216.121 (talk) 22:13, 1 November 2007 (UTC)

[edit] Important error

the limit of g(x) does not need to equal f(x)

--Perfection (talk) 17:38, 12 January 2008 (UTC)

[edit] explanation of revert

The edit by user:69.230.107.87 was good. The first algebraic manipulation is so clear that there is no need to spend a line doing it. The explanation stating that cos is continuous is precise, simple, correct and the established way to make such an argument. Writing cos(1/∞) is too informal. Oded (talk) 16:18, 1 May 2008 (UTC)

That was me (I apparently didn't notice I wasn't logged in at the time). It drives me crazy when people say or that . The limit as x approaches zero is infinity and the limit as x approaches infinity (meaning that x just gets bigger and bigger) is zero, but it is important to remember that it is impossible to actually "get" to zero or infinity, but we can get arbitrarily close.

miromodo (talk) 02:33, 6 May 2008 (UTC)

[edit] L'Hospital Proofs

Not sure about the proofs for L'Hospital's rule (forgive the mis-spelling), I changed the 0/0 case as using only epsilon as previously in both the numerator and denominator as this usage can be contradicted by the curves x^2 and x^3. I'm also not sure as to the veritability of the second proof, proofs I have seen in textbooks are far more complex and involve formal definitions of limits.
Cosec(x) (talk) 04:00, 20 May 2008 (UTC)

Could you clarify "can be contradicted by the curves x^2 and x^3"? Furthermore, your changes haven't helped. The final step:

is meaningless. By definition:

Therefore, I've reverted the changes again. Oli Filth^(talk) 19:52, 21 May 2008 (UTC)