Talk:L'Hôpital's Rule/Proof 2

Your comment "$\xi$ is not a function" may require an error page to be raised against the Binmore text -- I will take a look once I have sorted something out. --prime mover (talk) 18:18, 1 May 2014 (UTC)


 * Well, it's not like it's unsalvageable, but you need to use the axiom of choice to fix a function $\xi (x)$. This then has to be done explicitly. &mdash; Lord_Farin (talk) 21:38, 1 May 2014 (UTC)


 * I'm puzzled. There's nothing in Limit of Function in Interval that suggests that $\xi$ has to be a function of $x$. All it states is that if it is demonstrated that $f(x)$ is between $\xi$ and $x$ and $x \to \xi$, then $f(x) \to \xi$. Nothing about $\xi$ being a function (which was, as you say, a misleading statement made by Binmore. But it should be no more a problem than the $\epsilon$ - $\delta$ definition: all that is being stated is that it is possible to find a $\delta$ such that ...


 * The existence of such a $\xi$ is proven in Cauchy Mean Value Theorem. Then Limit of Function in Interval proves that the existence of such a $\xi$ proves that $f(x) \to \xi$. Where is the need for AoC? --prime mover (talk) 22:15, 1 May 2014 (UTC)


 * In Limit of Composite Function. &mdash; Lord_Farin (talk) 22:36, 1 May 2014 (UTC)


 * But that merely invokes the epsilon-delta definition of continuity: "Since $\ds \lim_{y \to \eta} f \left({y}\right) = l$, we can find $\Delta > 0$ such that: ..." . It is exactly the same construct that you are using in your Proof 1: "By the definition of limit, we ought to find a $\delta > 0$ such that: ..."


 * Does this highlight the fact that the definition of continuity is itself dependent on AoC? If so, then practically the whole of analysis needs to have the AoC template appended to the bottom. I may need to look at this more closely, because I'm pretty sure that I'm missing something important. --prime mover (talk) 22:47, 1 May 2014 (UTC)