Limit evaluation. In N. Piskunov he explained differential equations by taking up the example of air resistance acting on a falling body. After evaluating the differential equation he gets an equation for the velocity as:
$$v = \left(v_o - \frac{mg}{k}\right)e^{-\frac{kt}{m}} + \frac{mg}{k}.$$
He then states that if $k = 0$ then the equation turns to the basic equation:
$$v = v_o + gt.$$
Now I understand this statement because when air resistance is zero this velocity equation holds. However I am not able to prove this statement when I'm evaluating the limit:
$$\lim_{k \rightarrow 0} \left[\left(v_o - \frac{mg}{k}\right)e^{-\frac{kt}{m}} + \frac{mg}{k} \right]$$
Any help evaluating the limit would be appreciated!
 A: You have
$$\begin{equation}\begin{aligned}
\lim_{k \rightarrow 0} \left[\left(v_o - \frac{mg}{k}\right)e^{-\frac{kt}{m}} + \frac{mg}{k} \right] & = \lim_{k \rightarrow 0} \left[v_oe^{-\frac{kt}{m}} - \frac{mg}{k}e^{-\frac{kt}{m}} + \frac{mg}{k} \right] \\
& = \lim_{k \rightarrow 0} \left[v_{o}e^{-\frac{kt}{m}} + \frac{mg\left(1 - e^{-\frac{kt}{m}}\right)}{k} \right] \\
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
The limit of the first term, i.e., $v_{o}e^{-\frac{kt}{m}}$, is just simply $v_{o}$. For the second term, since the limit becomes it going to $\frac{0}{0}$, using L'Hôpital's rule, gives
$$\begin{equation}\begin{aligned}
\lim_{k \rightarrow 0} \frac{mg\left(1 - e^{-\frac{kt}{m}}\right)}{k} & = \lim_{k \rightarrow 0} \frac{mg\left(-e^{-\frac{kt}{m}}\left(-\frac{t}{m}\right)\right)}{1} \\
& = mg\left(\frac{t}{m}\right) \\
& = gt
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
You could also have used dnfu's question comment listed fact to get the same value. The combined result from \eqref{eq1A} thus gives
$$\lim_{k \rightarrow 0} \left[\left(v_o - \frac{mg}{k}\right)e^{-\frac{kt}{m}} + \frac{mg}{k} \right] = v_{o} + gt \tag{3}\label{eq3A}$$
A: As an alternative, we have that by standard limit $\frac{e^x-1}{x} \to 1$ as $x \to 0$
$$ \frac{mg\left(1 - e^{-\frac{kt}{m}}\right)}{k} =gt \,\frac{ e^{-\frac{kt}{m}}-1}{-\frac{kt}m} \to gt \cdot 1= g t$$
and the result follows.
A: With
$$e^{-\dfrac{kt}{m}}\approx 1-\dfrac{kt}{m},$$
$$\left(v_o - \frac{mg}{k}\right)e^{-\dfrac{kt}{m}} + \frac{mg}{k}$$
becomes
$$v_o+\left(g-\frac{kv_0}m\right)t\approx v_o+gt.$$
