Relationship between l'Hospital's rule and the least upper bound property.

Solution 1:

Here let's assume $F$ is non-archimedean.

For each infinitesimal* $x \in F^{\times}$, let $[x]$ denote the archimedean class $\bigcup \limits_{n \in \mathbb{N}^*} \left]-n |x|;-\frac{1}{n} |x|\right[ \cup \left]\frac{1}{n}\cdot|x|;n\cdot|x|\right[$ of $x$ in $F^{\times}$. Let $C$ denote the set of acrhimedean classes in $F^{\times}$.

($C$ is at least countable since for each infinitesimal $x$, $[x];[x^2];\ldots;[x^n];\ldots$ are distinct.)

Now let $\varphi$ be a choice function on $C$, and let $f$ be the map defined by $f(x) = \varphi([x])$ if $x$ is infinitesimal, and $f(y) = 0$ if $y$ is not an infinitesimal.

$f$ is differentiable on $F^{\times}$ with $f' = 0$ since for $x \neq 0$, $f$ is constant in the neighborhood $[x]$ of $x$. Moreover, $\lim \limits_{x \to 0} f(x) = 0$, since for $\varepsilon > 0$ infinitesimal, $\forall x \in \left]-\varepsilon^2;\varepsilon^2\right[$, $|f(x)| \leq \varepsilon$.

However, $\frac{f}{\operatorname{id}_{F^{\times}}}$ does not converge at $0$: given $r > 0$ and infinitesimal, and $n \in \mathbb{N}^*$, $[\frac{\varphi(r)}{n}] = [\varphi(r)]$, so $\dfrac{f(\frac{\varphi([r])}{n})}{\frac{\varphi([r])}{n}} = \dfrac{\varphi([r])}{\frac{\varphi([r])}{n}} = n$.

I believe this counterexample (inspired by that which you proposed for $\mathbb{Q}$), along with an adaptation of your argument for $\mathbb{Q}$ to any archimedean proper subfield of $\mathbb{R}$, proves that l'Hospital's rule is equivalent to the LUB property. (Incidently, the counterexample also fits my question since $f$ is not differentiable at $0$.)

*$x$ such that $\forall n \in \mathbb{N}, n |x| < 1$.