Talk:L'Hôpital's rule

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)

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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)

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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)

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)

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 $\lim _{x\to 0}{\frac {f(x)}{g(x)}}$ tends to 0/0, then $\lim _{x\to 0}f(x)=\lim _{x\to 0}g(x)=0$. 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 $f(x)\sim xf'(x)$ and $g(x)\sim xg'(x)$, both near $x=0$.

Hence $\lim _{x\to 0}{\frac {f(x)}{g(x)}}=\lim _{x\to 0}{\frac {xf'(x)}{xg'(x)}}=\lim _{x\to 0}{\frac {f'(x)}{g'(x)}}$.

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

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)

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.