How to take this limit?

Question

So I have an equation

$y(t) = \dfrac{1}{\alpha}(v_0sin(\theta) + \dfrac{g}{\alpha})(1-e^{-\alpha t}) - \dfrac{gt}{\alpha}$

And am supposed to show that if we take $\alpha$ to be small, then $y(t)$ reduces to

$y_2(t) = v_0sin(\theta)t - \dfrac{gt^2}{2}$

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First part of my issue:

I am able to get the factor of $t$ that it seems I need. In the limit of $\alpha \rightarrow 0$,

$\dfrac{1}{\alpha}(1 - e^{-\alpha t}) \rightarrow t$

I actually discovered this while messing around with some values, I can't find a way to actually prove it on paper by setting $\alpha$ to a small value. But I do think the claim is true.

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Second issue:

Other than that, I don't understand where the $\dfrac{gt^2}{2}$ comes from. Usingwhat I found above, I can simplify $y(t)$ to

$v_0 sin(\theta) t + \dfrac{gt}{\alpha} - \dfrac{gt}{\alpha} = v_0 sin(\theta) t$

But this obviously is wrong.

Answer 1

Proof of the stated limit: $$\lim_{\alpha \to 0} \frac{1 - e^{-\alpha t}}{\alpha} = -\lim_{\alpha \to 0} \frac{(e^{-t})^\alpha - (e^{-t})^0}{\alpha - 0}.$$ But by definition of the derivative $$f'(x) = \lim_{\alpha \to x} \frac{f(\alpha) - f(x)}{\alpha - x}$$ with the choice $x = 0$, $f(x) = (e^{-t})^x$, we observe the limit is simply $$\lim_{\alpha \to 0} \frac{1 - e^{-\alpha t}}{\alpha} = -\frac{d}{dx}\left[e^{-tx}\right]_{x = 0} = \left[te^{-tx}\right]_{x=0} = t.$$

I should have been more careful with the second part of your question, which involves the calculation of a distinct limit from the one above:

$$\tag{$*$}\lim_{\alpha \to 0} \frac{1 - e^{-\alpha t}}{\alpha^2} - \frac{t}{\alpha}.$$ The reason for this is because we cannot in general write $$\lim_{x \to 0} f(x)g(x) = \lim_{x \to 0} f(x) \cdot \lim_{x \to 0} g(x).$$ This much was pointed out by another answer that is as of this writing, deleted.

To evaluate this limit, we can employ the series expansion of $e^{-\alpha t}$ around $\alpha = 0$: $$e^{-\alpha t} = \sum_{k=0}^\infty \frac{(-\alpha t)^k}{k!} \sim 1 - \alpha t + \frac{(\alpha t)^2}{2} - O((\alpha t)^3).$$ Thus for fixed $t$, $$\frac{1 - e^{-\alpha t}}{\alpha^2} - \frac{t}{\alpha} \sim -\frac{t}{\alpha} + \frac{\alpha t - (\alpha t)^2/2}{\alpha^2} + O(\alpha) \sim - \frac{t^2}{2} + O(\alpha),$$ and as $\alpha \to 0$, we obtain the desired behavior. We can also use L'Hopital's rule to evaluate the limit $(*)$, or other methods. Perhaps, suitably inspired by the use of the definition of derivative to evaluate the first limit, you might be able to formulate a similar argument for the second.

Answer 2

Considering $$y=\frac{ \left(A+\frac{g}{\alpha }\right)\left(1-e^{-\alpha t}\right)}{\alpha }-\frac{g t}{\alpha }$$ Use Taylor expansion around $\alpha=0$ $$e^{-\alpha t}=1-\alpha t+\frac{\alpha ^2 t^2}{2}+O\left(\alpha ^3\right)$$ which makes $$\frac{1-e^{-\alpha t}}{\alpha }=t-\frac{\alpha t^2}{2}+O\left(\alpha ^2\right)$$ Then $$y=\left(A+\frac{g}{\alpha }\right)\left(t-\frac{\alpha t^2}{2}+O\left(\alpha ^2\right)\right)-\frac{g t}{\alpha }=\left(A t-\frac{g t^2}{2}\right)+O\left(\alpha \right)$$ Using one more term for the Taylor expansion, you could arrive to $$y=\left(A t-\frac{g t^2}{2}\right)+\frac{1}{6} \alpha \left(g t^3-3 A t^2\right)+O\left(\alpha ^2\right)$$ which shows the limit and how it is approached.