Let $(k,+,.,0,1,<)$ be an ordered field. In the folowing definitions, all numbers and notions are derived from the ordered field structure of $k$, and $a < c$ are generic elements of $k$.
Define the limit of a function $f: ]a;b[ \cup ]b;c[ \longrightarrow k$ at $b$ as the element $l$ of $k$ if there is such element which satisfies $\forall \varepsilon > 0, \exists \eta > 0, \forall x \in ]a;b[ \cup ]b;c[, |x-c| < \eta \rightarrow |f(x) - l| < \varepsilon$. The limit if it exists is unique. (this coïncides with the synthetic definition of a limit if we equip $k$ with its order topology)
Let's say that a map $f: ]a;c[ \longrightarrow k$ is continuous at $b \in ]a;c[$ iff $f(b)$ is the limit of $f|_{]a;b[ \cup ]b;c[}$ at $b$. Let's say that a map $f: ]a;c[ \longrightarrow k$ is differentiable at $b \in ]a;c[$ iff the map $g: ]a;b[ \cup ]b;c[ \longrightarrow k$ defined by $g(x) = \frac{f(x) - f(b)}{x-b}$ has a limit. Then let $f'(b)$ denote this limit.
I wonder if the following theorem of real analysis holds only if $k$ is isomorphic to $\mathbb{R}$:
Let $f: ]a;c[ \longrightarrow k$ be continuous at every point in $]a;c[$, and let's assume there is $b \in ]a;c[$ such that $f$ differentiable at every point of $]a;b[ \cup ]b;c[$. If $x \in ]a;b[ \cup ]b;c[ \mapsto f'(x)$ has a limit at $b$, then $f$ is differentiable at $b$ and $f'(b)$ is that limit.
The classical proof in $\mathbb{R}$ uses the Intermediate Value Theorem (which implies the Least Upper Bound Porperty) but I am not that sure it is equivalent to it, the moral reason being that this theorem isn't really a theorem of existence. Now "$\forall f: [a;b] \rightarrow k$ continuous, if $f$ is injective then it is stricly monotone" isn't really a theorem of existence either but it implies LUB so this doesn't mean anything.
Do you know if the above metionned theorem implies LUB for $k$?
This question is answered by the positive
-in the answer to this topic for the non archimedean case: Relationship between l'Hospital's rule and the least upper bound property.
-in the counterexample cited in the of question of the same topic, though one has to adapt the proof a little bit (setting $f(x) = 2^{-n}$ instead of $f(x) = 8^{-n}$ for instance) so that $\lim \limits_{x \to 0} \frac{f(x)}{x} = +\infty$ while $\lim \limits_{x \to 0} \frac{f'(x)}{1} = 0$.