Answer
(a) $\lim\limits_{t \to \infty}v = \frac{mg}{c}$
This limit is the terminal velocity of the object as it falls.
(b) $\lim\limits_{c \to 0^+}v = gt$
The velocity of a falling object in a vacuum continues increasing as time passes.
Work Step by Step
(a) $\lim\limits_{t \to \infty}v$
$=\lim\limits_{t \to \infty}\frac{mg}{c}(1-e^{-ct/m})$
$=\lim\limits_{t \to \infty}\frac{mg}{c}(1-\frac{1}{e^{ct/m}})$
$=\frac{mg}{c}(1-0)$
$=\frac{mg}{c}$
This limit is the terminal velocity of the object as it falls.
(b) $\lim\limits_{c \to 0^+}v$
$=\lim\limits_{c \to 0^+}\frac{mg}{c}(1-e^{-ct/m})$
$=\lim\limits_{c \to 0^+}\frac{mg-mge^{-ct/m}}{c} = \frac{0}{0}$
We can use L'Hospital's Rule:
$\lim\limits_{c \to 0^+}\frac{mg-mge^{-ct/m}}{c}$
$=\lim\limits_{c \to 0^+}\frac{(-\frac{t}{m})(-mge^{-ct/m}~)}{1}$
$=\lim\limits_{c \to 0^+}(\frac{t}{m})(mge^{-ct/m})$
$=\lim\limits_{c \to 0^+}gt~e^{-ct/m}$
$=gt$
The velocity of a falling object in a vacuum continues increasing as time passes.
