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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!

Siddhanth Iyengar
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3 Answers3

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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

$$\lim_{k \rightarrow 0} \left[\left(v_o - \frac{mg}{k}\right)e^{-\frac{kt}{m}} + \frac{mg}{k} \right] = v_{o} + gt \tag{3}\label{eq3A}$$

John Omielan
  • 47,976
  • This is circular reasoning, you compute the limit of the difference quotient via l'Hôpital, which uses derivatives, which are defined as limits of difference quotients. It is not unusual to make such a complicated reasoning, but mathematics is also distilling down the result to the bare necessities,... – Lutz Lehmann Sep 22 '20 at 16:15
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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.

user
  • 154,566
1

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

$$v_o+\left(g-\frac{kv_0}m\right)t\approx v_o+gt.$$