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<r<1} |f(r)| < \infty, 
\end{equation*}
and
\begin{equation*}
\liminf_{r \to 0+} \,f'(r) \geq 0.
\end{equation*}
Can we conclude that $f(0+)$ exists? 
 A: Hint: Let $g(r)= r+f(r).$ Then $g'(r)> 0$ for $r$ in some $(0,a).$
A: Your first condition says $f$ is bounded on $(0,1)$.
Case I:    $\underset{r \rightarrow 0+}{\text{lim inf}} f'(r)>0$
In this case the lim inf condition says that eventually, for some $\epsilon>0$ 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:    $\underset{r \rightarrow 0+}{\text{lim inf }} f'(r)=0$
Put $g(r):=r+f(r)$
So $g'(r)=1+f'(r) \Rightarrow \underset{r \rightarrow 0+}{\text{lim inf }} g'(r)= 1 + \underset{r \rightarrow 0+}{\text{lim inf }} f'(r)=1>0$
Again, the lim inf condition says that eventually, for some $a>0$ sufficiently small, $g'$ is positive on $(0,a)$, which means $g$ is increasing on $(0,a)$.
Note $g$ is also bounded on $(0,1)$.
$g$ bounded and increasing on $(0,a)$ implies $g(0+):= \underset{r\rightarrow 0+}{\text{lim }} g(r) = \underset{r\in(0,a)}{\inf}g(r) \in \mathbb{R}$.
But also $g(0+):= \underset{r\rightarrow 0+}{\text{lim }} g(r) =  \underset{r\rightarrow 0+}{\text{lim }} (r+f(r))- \underset{r\rightarrow 0+}{\text{lim }} r= \underset{r\rightarrow 0+}{\text{lim }} f(r) = f(0+)$.
A: Your first condition says $f$ is bounded on $(0,1)$.
Case I:    $\underset{r \rightarrow 0+}{\text{lim inf}} f'(r)>0$
In this case the lim inf condition says that eventually, for some $\epsilon>0$ 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.)
