Answer
$P=\frac{C}{e^{c_1}e^{-rt}}$
$\underset{t \rightarrow \infty}{\lim} P(t)=C$
Work Step by Step
Follow the Gumpertz Population Model:
$$\frac{dP}{dt}=rP(\ln C-\ln P)$$
$$r=\frac{dP}{dt}\frac{1}{P(\ln C-\ln P)}$$
$$$rdt=dP\frac{1}{P(\ln C-\ln P)}$$
Intergrate:
$$\int rdt=\int \frac{1}{P(\ln C-\ln P)}dP$$
$$-\ln(\ln C- \ln P)=rt+c_1$$
Solve for P:
$$\ln(\ln C- \ln P)=-rt-c_1$$
$$\ln C- \ln P=c_1e^{rt}$$
$$\ln (\frac{C}{P})=c_1e^{rt}$$
$$\frac{C}{P}=e^{c_1}e^{rt}$$
$$P=\frac{C}{e^{c_1}e^{-rt}}$$
Find the limit:
$$\underset{t \rightarrow \infty}{\lim} e^{-rt}=0$$
and $$\underset{t \rightarrow \infty}{\lim} e^{c_1(0)}=1$$
Hence, $\underset{t \rightarrow \infty}{\lim} P(t)=C$
