Given
$r(t)=\frac{f(t)}{1-F(t)} \tag{Eq. 1}$ where $f(t)=\frac{dF}{dt} \tag{Eq. 2}$
and the conditions:
$\lim_{t\rightarrow \infty} r(t)=1 \tag{Eq. 3}$ $\lim_{t\rightarrow \infty} F(t)=1 \tag{Eq. 4}$ $\lim_{t\rightarrow \infty} f(t)=1-F(t) \tag{Eq. 5}$
I can think of just one function $F$ satisfying these three conditions--the logistic function:
$F(t)=\frac{1}{1+e^{-t}} \tag{Eq. 6}$
(which can also be expressed $F(t)=r(t)$)
Is this is the only function satisfying these conditions? If so, is there a way to prove it?