Let $E$ be a (separated) topological vector space over $\mathbb{R}$, $f\colon [0,1]\to E$ continuous. Assume that for every $t \in (0,1)$ we have a derivative $$f'(t) = \lim_{h\to 0} \frac{f(t+h)- f(h)}{h}$$
Moreover, assume that there exists the limit $\lim_{t \to 0_{+}} f'(t)$
Is it true that $$\lim_{h\to 0_{+}} \frac{f(h)-f(0)}{h}=\lim_{t \to 0_{+}} f'(t) $$ ?
Notes:
This is true for $E$ finite dimensional topological vector space ( since it's isomorphic to $\mathbb{R}^n$, and we can work component-wise, using Lagrange intermediate value theorem).
If $E$ is arbitrary, we can still conclude $$ \frac{f(h) - f(0)}{h} \to \lim_{t\to 0} f'(t)$$ weakly.
If $f'$ has a continuous extension to $[0,1]$, and $E$ is locally convex and complete ( say a Banach space) then we have $$f(h) - f(0) = \int_{0}^h f'(t) dt = h \cdot \int_{0}^1 f'(h t) d t$$ so again we get the desired equality.
It may be that the equality ( L'Hospital rule) is not true if $E$ is not locally convex. It may have to do with the fact that the average of a sequence convergent to $0$ may not converge to $0$. Or it may not work in the general form even for locally convex spaces, or even Banach spaces.
$\bf{Added:}$ I think it will work for locally convex spaces. We use a bit of compactness too. But I haven't written down details. Anyways, maybe it's a very basic fact, for which references are available.
$\bf{Added:}$ The following lemma is useful, and not hard to prove (use the compactness of $[a,b]$):
Let $[a,b]$ an interval, $f\colon [a,b]\to E$, with a derivative at every point. Let $U$ be an open convex set such that $f'(t) \in U$ for all $t \in [a,b]$. Then $$\frac{f(b)-f(a)}{b-a} \in U$$
$\bf{Added:}$ Another lemma : Let $[a,b]$ an interval, $f\colon [a,b]\to E$, with a derivative at every point except a finite set $A$. Let $U$ be an open convex set such that $f'(t) \in U$ for all $t \in [a,b]\backslash A$. Then
$$\frac{f(b)-f(a)}{b-a} \in \bar U$$
( similar to Dieudonne lemma here... perhaps it's also valid with the proper modifications).
Note: this will take care of the case $E$ locally convex.
$\bf{Added:}$ It turns out that the statement is related to the mean value theorem "inequality", as seen from the previous lemma. What I have learned and is somehow related is that the inequality might not be true if the topological vector space $E$ does not have any linear functionals $\ne 0$. There exist functions $f\colon [0,1]\to E$, $f'(t) = 0$ for all $t \in [0,1]$, but $f$ not constant. See " functions with $0$ derivative".
$\bf{Added:}$ This is the example of Rolewicz of a nonconstant function with $0$ derivative from $[0,1]$ to $L^p[0,1]$ ( $0<p<1$). Take $f(t) = \chi_{[0,t]}$ (!). Following this idea, here is an example of a continuous function $f\colon [0,1] \to L^p[0,1]$ such that $f'(t) = 0$ for all $t \in (0,1]$, but $\frac{f(h)-f(0)}{h}$ does not converge to $0$ as $h\to 0_{+}$. Just take $$f(t) = \chi_{[0, t^p]}$$
This provides a counterexample.
$\bf{Added:}$ Now that we know it may not work for spaces what are not locally convex, let's see what more can be said.
With the same methods we can show: if $f\colon [0,1]\to E$ , $g\colon [0,1] \to \mathbb{R}$ continuous functions, with derivatives on $(0,1]$, and moreover: $E$ is locally convex and $g$ is monotonous around $0$ (I cannot avoid this extra hypothesis for general l.c. $E$ ), and moreover $\lim_{t\to 0_{+}} \frac{f'(t)}{g'(t)}$ exists then $$\lim_{t\to 0_{+}} \frac {f(t)-f(0)}{g(t)-g(0)} = \lim_{t\to 0_+} \frac{f'(t)}{g'(t)}$$
