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}

\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)