Does $\liminf_{r\to 0+} f'(r) \geq 0$ implies the existence of $f(0 +)$? Let $f$ be a real-valued differentiable function on $(0,1)$, and suppose that
\begin{equation*}
\sup_{0&lt;r&lt;1} |f(r)| &lt; \infty, 
\end{equation*}
and
\begin{equation*}
\liminf_{r \to 0+} \,f'(r) \geq 0.
\end{equation*}
Can we conclude that $f(0+)$ exists?
Hint: Let $g(r)= r+f(r).$ Then $g'(r)&gt; 0$ for $r$ in some $(0,a).$
Your first condition says $f$ is bounded on $(0,1)$.

Case I:    $\underset{r \rightarrow 0+}{\text{lim inf}} f'(r)&gt;0$

In this case the lim inf condition says that eventually, for some $\epsilon&gt;0$ sufficiently small, $f'$ is positive on $(0,\epsilon)$, which means $f$ is increasing on $(0,\epsilon)$.

$f$ bounded and increasing on $(0,\epsilon)$ implies $f(0+):= \underset{r\rightarrow 0+}{\text{lim }} f(r) = \underset{r\in(0,\epsilon)}{\inf}f(r) \in \mathbb{R}$.

Case II:    $\underset{r \rightarrow 0+}{\text{lim inf}} f'(r)=0$

...

(It is not clear to me whether $f(0+)$ exists in this case.)