Answer
$$\frac{R}{r}$$
Work Step by Step
Since we have for $ R(t) =R$ $$
P V = \frac{R}{r}\left(1-e^{-r T}\right)
$$
Then for large $T$
\begin{align*}
PV&=\lim_{T\to \infty }\frac{R}{r}\left(1-e^{-r T}\right)\\
&=\frac{R}{r}
\end{align*}
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