Exchange of the one-sided derivative and the one-sided limit of a real variable Let $f$ be a real-valued continuous function on $[0, 1)$. Suppose that $f$ is infinitely-differentiable on $(0 , 1)$. Is there some sufficient condition for $f$ to be right-differentiable at $0$ with
$$f_{+}'(0) = \lim_{x \to 0^{+}} f'(x),$$
where I use the notation $f_{+}'(0) := \lim_{\Delta x \to 0^+} \frac{f(\Delta x) - f(0)}{\Delta x}$ for the right derivative of $f$ at $0$?

If a function is uniformly continuous on the open interval $(a,b)$ then it can be extended to a continuous function on $[a,b]$; by defining $f(a)=\lim_{x\rightarrow a^+} f(x)$ and $f(b)=\lim_{x\rightarrow b^-} f(x)$. So if your first derivative is uniformly continuous on $(0,1)$ then both limits $f^{\prime}_{+}(0)=\lim_{x\rightarrow 0^+} f^{\prime}(x)$ and $f^{\prime}_{-}(1)=\lim_{x\rightarrow 1^-} f^{\prime}(x)$ exist and give a continuous extension of $f^{\prime}$ to $[0,1]$.
