Talk:L'Hôpital's Rule

first discussion
Let's talk before we fill up the history of this page with a big ol' discussion.

I contend that the limit of f'/g' will exist, so long as the derivatives exist. is this not right?

also, i think that my previous list of categories was overkill (i do that sometimes), but "Limits" is a legitimate category, since they are a part of calculus, AND analysis --Jehan60188 16:55, 4 October 2008 (UTC)

"Limits" is good for me as a category.

I'm racking my memory from many-years-ago study, but I believe that the limit of f/g = limit of f'/'g as long as the limit of f'/g' exists. In practice you often get f'/g' = 0. No problem, just diff again: the limit of f'/g' = limit of f"/g" and so on up the drivs till you get a pair that make sense.

Feel free to get cracking on the calculus, I'm still bogged down in defining the number systems. --prime mover (talk) 17:01, 4 October 2008 (UTC)

The limit does not have to exist. Take for instance $f(x)=x+sin(x)$ and take $g(x)=x$. Then the derivatives of $f$ and $g$ will exist, but the limit as $x\rightarrow\infty$ of $\frac{f'(x)}{g'(x)}$ will not exist.

As for the categories, I agree, but make sure you separate categories with a ';'. It might also be nice to mention the indeterminate form of the limit and show the other types, which will all reduce to either $\frac{0}{0}$ or $\frac{\infty}{\infty}$. --Joe 17:05, 4 October 2008 (UTC)

Wikipedia and Mathworld both agree that lim f'/g' has to exist, so I'm adding that to the page. -RickettsAM 14:16, 5 October 2008 (UTC)

Revert
I reverted not because the statement was wrong, but because this would introduce a crippling disparity between the statement and the proof. Both should be addressed at the same time. Probably best to delay it though, as the PW coverage of infinite limits is very sparse at the moment. &mdash; Lord_Farin (talk) 19:03, 18 January 2014 (UTC)


 * It is of course easy to add the $\pm \infty$ case into a corollary and return to the existing result by taking reciprocals of $f$ and $g$. This is effectively what we did in college. --prime mover (talk) 19:42, 18 January 2014 (UTC)

Change of proof
The proof had to be changed due to its imprecision. The premise that $\xi$ were a function of $x$ was not correct -- many $\xi$ could correspond to one $x$. The least ramification would be an invocation of AC to choose a $\xi$ for every $x$.

However, the proof is really simple in the language of $\epsilon$-$\delta$, so I went for that instead. I hope you don't mind. &mdash; Lord_Farin (talk) 11:02, 30 April 2014 (UTC)


 * The original proof was apparently copypasta'd from Wikipedia and dates from 2008. I'm easy with what you have done. --prime mover (talk) 12:09, 30 April 2014 (UTC)


 * Sorry about the "is a function of $x$", I wasn't thinking -- I was just trying to clear up the waffle around how $\xi$ depended upon $x$, should have deleted that line completely as it was superfluous. I may reinstate the proof as I've now got round to checking it in Binmore. --prime mover (talk) 12:12, 30 April 2014 (UTC)


 * We now have 2 proofs. I rewrote the line about "function of $x$" to be more user-friendly. --prime mover (talk) 08:31, 1 May 2014 (UTC)