How to prove $\lim_{k\rightarrow0}\frac{2t}{k}-\frac{2}{k^2}\left(1-e^{-kt}\right)=t^2$ When playing around with free fall with and without air resistance, I noticed something funny:
If we assume no air resistance, constant gravity $g$, velocity $v_0$ and position $y_0$ at $t=0$, we can set up $\Sigma F=ma=mg$ and get the classical formulas by integration:
Acceleration: $a_A(t)=g$
Velocity: $v_A(t)=gt+v_0$
Position: $y_A(t)=\frac12gt^2+v_0t+y_0$
If we, however, assume air resistance to be proportional to, and negative of velocity, we can set up $\Sigma F=ma=mg-kv$ and get the following (with a bit more work):
Acceleration: $a_B(t)=\left(g-\frac{k}{m}v_0\right)e^{-\frac{k}{m}t}$
Velocity: $v_B(t)=\frac{m}{k}g+\left(v_0-\frac{m}{k}g\right)e^{-\frac{k}{m}t}$
Position: $y_B(t)=y_0+\frac{m}{k}\left(gt-\frac{m}{k}\left(g-\frac{k}{m}v_0\right)\left(1-e^{-\frac{k}{m}t}\right)\right)$
Here I became curious about the functions $y_A(t)$ and $y_B(t)$, and what would happen if $k$ becomes very small. This implies that air resistance becomes small, and that $y_B(t)$ starts behaving like $y_A(t)$. But I found that counter-intuitive because $y_A(t)$ is basically a geometric function, while $y_B(t)$ is somewhat exponential.
So I compared them graphically, setting $y_0=v_0=0$, $m=1$, $g=2$, and $k=\textrm{very small}$, and, indeed, the graphs merge for low values of $t$. With these values, $y_A(t)$ becomes $t^2$, and the expression for $y_B(t)$ becomes $\frac{2t}{k}-\frac{2}{k^2}\left(1-e^{-kt}\right)$. This made me convinced that, with these values for $y_0, v_0, m$ and $g$, that $\lim_{k\rightarrow0}y_B(t)=y_A(t)$, that is:
$$\lim_{k\rightarrow0}\frac{2t}{k}-\frac{2}{k^2}\left(1-e^{-kt}\right)=t^2$$
My question is: How do we prove this limit (if it is correct)?
 A: You can have a better understanding of this problem after learning Taylor series and Laurent series. We use the Taylor series expansion
$$e^{-kt}=1-kt+\frac{k^2t^2}{2!}-\frac{k^3t^3}{3!}+\cdots.$$
The $k^2$ in the denominator forces us to go to high orders to get correct results. So the function
\begin{align}
&\quad\;\frac{2t}{k}-\frac{2}{k^2}(1-e^{-kt})=\frac{2t}{k}-\frac{2}{k^2}\left(kt-\frac{k^2t^2}{2}+\frac{k^3t^3}{6}-\cdots\right)\\
&=\frac{2t}{k}-\frac{2t}{k}+t^2-\frac{kt^3}{6}+\cdots=t^2-\frac{kt^3}{6}+\cdots.
\end{align}
In the limit $k\rightarrow0$, the divergent Laurent terms all cancel out and the function approaches $t^2$ with infinitesimal higher-order correction terms. But the convergence is point-wise, meaning no matter how small the resistance $\,k\,$ is, in the long run the motion would be very different from a free fall. So physically speaking, the free-fall approximation is good for $kt\ll 1$.
A: By twice using L'Hopital's Rule below, we have:
$$
\lim_{k\to 0}\left(\frac{2t}{k}-\frac{2}{k^2}(1-e^{-kt})\right)=\lim_{k\to 0}\frac{2tk-2(1-e^{-kt})}{k^2}=\cdots=\lim_{k\to 0} e^{-kt}t^2=t^2.
$$
