Talk:L'Hôpital's Rule/Also defined as

There is no requirement that $\map f x \to \infty$ as $x \to b$, so the function cannot be proven discontinuous there. In fact, we could take $b' = \dfrac {a + b} 2$, and the theorem would apply on $\hointl a {b'}$, but $f$ would necessarily be continuous at $b'$. --CircuitCraft (talk) 18:44, 25 September 2023 (UTC)


 * That's not the point. You can have it dicsontinuous at either end, or both. --prime mover (talk) 19:26, 25 September 2023 (UTC)


 * You can have discontinuities on either end, but the assertion is that $f$ and $g$ must have discontinuities on $a$ and $b$ (in Corollary 2). This is not true. They may have a discontinuity at $b$, but they must have a discontinuity at $a$. --CircuitCraft (talk) 19:29, 25 September 2023 (UTC)


 * Clarified the language. --prime mover (talk) 19:45, 25 September 2023 (UTC)


 * Still not right. There isn't any symmetry between $a$ and $b$. In fact, we only every care about $a$ when applying L'Hopital's Rule; $b$ just needs to exist so that we have an interval to do proofs on. Think of it like an open neighborhood of $a$, but only on one side.
 * What I'm trying to point out is that, in the statement of L'Hôpital's Rule/Corollary 2, we assume:
 * $\map f x \to \infty$ and $\map g x \to \infty$ as $x \to a^+$
 * Note that there is no equivalent assumption for $b$. It's that assumption that forces a discontinuity and/or not being defined at that point. There's an equivalent version of the rule that works upwards instead, but that's not the one we have listed. (It also follows trivially from what we do have). --CircuitCraft (talk) 22:08, 25 September 2023 (UTC)