Talk:L'Hôpital's Rule/Corollary 2
Doesn't the fact that $f$ and $g$ are continuous on $\closedint a b$ contradict the fact that they diverge at $a$? I think the statement here might be flawed. --CircuitCraft (talk) 02:54, 25 September 2023 (UTC)
- Abramowitz and Stegun (the only source for this specific version of this result I have found) just defines differentiability on $\openint a b$. I can't see why continuity on that interval was specified for the main result, even, but Binmore (cited in Proof 2) specifies it, as it's specified for the Cauchy mean value theorem (which he uses to prove it). I would have thought that differentiability on $\openint a b$ should be enough for anybody. Requiring continuity at the endpoints seems to be overegging it. --prime mover (talk) 05:08, 25 September 2023 (UTC)
- I think that covers it. --prime mover (talk) 16:36, 25 September 2023 (UTC)
- Requiring continuity on $\closedint a b$ is a simple way to state the "real" requirement that $\ds \lim_{x \to a^+} \map f x = 0$ and likewise for $g$. If we have that, we can always "fill in" the missing point as:
- $\map {f_0} x = \begin{cases}
- Requiring continuity on $\closedint a b$ is a simple way to state the "real" requirement that $\ds \lim_{x \to a^+} \map f x = 0$ and likewise for $g$. If we have that, we can always "fill in" the missing point as:
- I think that covers it. --prime mover (talk) 16:36, 25 September 2023 (UTC)
\map f x & : x \in \openint a b \\ 0 & : x = a \end{cases}$
- on which we can apply the Cauchy Mean Value Theorem. This is not required in L'Hôpital's Rule/Corollary 2, since we only apply that theorem on $\closedint {x_{n - 1}} {x_n} \subsetneq \openint a b$, on which the function is continuous and differentiable anyway. --CircuitCraft (talk) 16:58, 25 September 2023 (UTC)
- except they tend to infinity not zero --prime mover (talk) 18:33, 25 September 2023 (UTC)
- I was referring to the main result, where they do tend to zero. There is a good reason Binmore required it; that's because it is in fact required, provided we write $\map f a = 0$. By writing $\ds \lim_{x \to a^+} \map f x = 0$, we bypass that requirement. --CircuitCraft (talk) 18:46, 25 September 2023 (UTC)
- Might be worth taking this discussion to that page then, sorry, I thought we were talking about this one --prime mover (talk) 20:59, 26 September 2023 (UTC)
Sorry, I'm re-reading all this and yes I appreciate there are complexities here. Thought the discussion had been about language used.
Yes I appreciate that we probably need to branch that "also defined as" page out even further and say "If we assume continuity at endpoint, then we don't have to make that limiting argument, and so ..."
Can we extract this very point into a lemma? It could say what needed to be filled in before you apply Cauchy. Sorry, exhausted tonight. --prime mover (talk) 20:57, 26 September 2023 (UTC)