What happens when time approaches infinity in differential equation $m\frac{dv}{dt} = mg - kv$. We know that the differential equation for velocity is $m \, dv/dt = mg - kv$
where $k$ is air drag. What I am wondering is what happens as time approaches infinity. How is terminal velocity expressed in terms of this equation?
 A: When $mg=kv,$ then the right side of your differential equation is $0.$ That means $dv/dt=0$ so the velocity doesn't change. Solving $mg= kv$ for $v$ gives you terminal velocity, and that is the limiting velocity as $t\to\infty.$
A: Let $k\geq0$ then
\begin{eqnarray*}
mg-kv              &=& ma \\
ma+kv              &=& mg ~~~~~~;~~~~~~a=v'\\
v'+\frac{k}{m}v    &=& g    
\end{eqnarray*}
which is differential equation and with integration factor $\displaystyle I=e^{\frac{k}{m}t}$ is
$$e^{\frac{k}{m}t}v=\int e^{\frac{k}{m}t}g~dt+C=\frac{mg}{k}e^{\frac{k}{m}t}+C$$
hence $\displaystyle v=\frac{mg}{k}+Ce^{-\frac{k}{m}t}$ is velocity equation. If falling be in free thus $v_0=0$ and $\displaystyle C=-\frac{mg}{k}$, so
$$\displaystyle v=\frac{mg}{k}(1-e^{-\frac{k}{m}t})$$
is velocity equation. It's graph shows the velocity doesn't acceed a value $\displaystyle v_T=\frac{mg}{k}$ because
$$|v|=\Big|\frac{mg}{k}\big(1-e^{-\frac{k}{m}t}\big)\Big|\leq\frac{mg}{k}=v_T$$
and this value $v_T$ is Terminal Velocity. Also
$$\displaystyle\lim_{t\to\infty}v=\lim_{t\to\infty}\frac{mg}{k}(1-e^{-\frac{k}{m}t})=v_T$$
this subject is true when an object falls through a fluid, like raindrop.
