Answer
(a) $\lim\limits_{t \to \infty}P(t) = M$
This answer is to be expected. The text in the question states: "$M$ is called
the carrying capacity and represents the maximum population size that can be supported."
(b) $\lim\limits_{M \to \infty}P(t) = P_0~e^{kt}$
Note that this function is an exponential function.
Work Step by Step
(a) $\lim\limits_{t \to \infty}P(t) = \lim\limits_{t \to \infty}\frac{M}{1+Ae^{-kt}}$
$\lim\limits_{t \to \infty}P(t) = \frac{M}{1+0}$
$\lim\limits_{t \to \infty}P(t) = M$
This answer is to be expected. The text in the question states: "$M$ is called
the carrying capacity and represents the maximum population size that can be supported."
(b) $\lim\limits_{M \to \infty}P(t) = \lim\limits_{M \to \infty}\frac{M}{1+Ae^{-kt}}$
$\lim\limits_{M \to \infty}P(t) = \lim\limits_{M \to \infty}\frac{M}{1+\frac{M-P_0}{P_0e^{kt}}}$
$\lim\limits_{M \to \infty}P(t) = \lim\limits_{M \to \infty}\frac{M}{\frac{P_0~e^{kt}+M-P_0}{P_0e^{kt}}}$
$\lim\limits_{M \to \infty}P(t) = \lim\limits_{M \to \infty}\frac{M~P_0~e^{kt}}{P_0~e^{kt}+M-P_0}$
$\lim\limits_{M \to \infty}P(t) = \frac{\infty}{\infty}$
We can apply L'Hospital's Rule:
$\lim\limits_{M \to \infty}P(t) = \lim\limits_{M \to \infty}\frac{M~P_0~e^{kt}}{P_0~e^{kt}+M-P_0}$
$\lim\limits_{M \to \infty}P(t) = \lim\limits_{M \to \infty}\frac{P_0~e^{kt}}{0+1+0}$
$\lim\limits_{M \to \infty}P(t) = P_0~e^{kt}$
Note that this function is an exponential function.
