Talk:Calculus/L'Hôpital's Rule

Shouldn't L'Hôpital's rule go in application of differentiation header since if this was being used to learn Calculus you wouldn't know what a derivative is yet. If this book however is meant as a reference book I understand the placement.
 * I agree. +1 for moving L'Hôpital's rule to a later section.  Greenbreen (discuss • contribs) 16:31, 23 March 2011 (UTC)
 * Done --Greenbreen (discuss • contribs)

Proofs
I added the proof of the 0/0 case. I looked at the proof of the $$\infty/\infty$$ case, and it seems considerably more complicated. I'm not sure if I should have added the proof of just one case, but I'll leave it here for now and see how it goes. --Greenbreen (discuss • contribs) 18:12, 22 May 2011 (UTC)

Removed alternate proof
I removed the following alternate "proof" of L'Hôpital's rule for the 0/0 case:

I think the argument suffers from multiple issues, but here is a specific counter-example to show that the proof does not hold up. Let $$f(x)=g(x)=x^2$$, and let $$a=0$$. Then $$f$$ and $$g$$ are continuous and differentiable everywhere and $$f(a)=g(a)=0$$. The limit
 * $$\lim_{h\to 0}\frac{f(a+h)}{g(a+h)}$$

exists and is equal to $$1$$. However,
 * $$\lim_{h\to 0}\frac{\frac{f(a+h)-f(a)}{h}}{\frac{g(a+h)-g(a)}{h}}=\frac{f'(a)}{g'(a)}$$

is not defined because the denominator is 0.

Of course the real problem is that what is shown is not L'Hôpital's rule at all since it never mentions a limit that involves $$x$$. --Greenbreen (discuss • contribs) 19:32, 27 September 2011 (UTC)