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Here is my problem. I reproduce the two questions below. I'm not satisfied at all with how I solved this. Are there any errors?

$$\begin{split} I(t)& = \frac{2A}{w_{0}^2-w^2}\sin\frac{(w_{0}-w)t}{2}\sin\frac{(w_{0}+w)t}{2}\\ I_{0}(t)&=\lim_{w\to w_0}I(t). \end{split}$$

$A$ and $w_{0}$ are constant. $t$ is fixed for the first part.

The first part is just asking you to find $I_{0}(t)$.

I started off using $$\sin(a)\sin(b) = \frac{1}{2}(\cos(a-b) - \cos(a+b)),$$

$$ \frac{(w_{0}-w)t}{2}-\frac{(w_{0}+w)t}{2} = -wt,$$ and $$ \frac{(w_{0}-w)t}{2}+\frac{(w_{0}+w)t}{2} = w_{0}t,$$ so, $$\lim_{w \to w_{0}}\frac{A}{w_{0}^2-w^2}(\cos(-wt) - \cos(w_{0}t)) $$

Then I used l'Hospital's rule and plugged in the limit afterwards to get $$\frac{At\sin(-w_{0}t)+A\sin(w_{0}t)}{-2w_{0}}$$

I'm very unsatisfied with this result, but I don't know where I went wrong if I did.

The second part of the question is also something I'm not sure how to do. It's asking you to show that $I(t)$ is bounded, but $I_{0}(t)$ isn't. I just wrote this explanation out:

$I(t)$ is bounded because $\sin\frac{(w_{0}-w)t}{2}$ and $\sin\frac{(w_{0}+w)t}{2}$ are bounded between $1$ and $-1$, so $I(t)$ is bounded between $\frac{2A}{w_{0}^2-w^2}$ and $\frac{-2A}{w_{0}^2-w^2}$. However, $At\sin(-w_{0}t)$ is unbounded as it can go to $\infty$ or $-\infty$, so the entire function $I_{0}(t)$ is unbounded.

Anne Bauval
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Alexander Aseme
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    It might be a good idea to write a more descriptive title, so users with relevant knowledge can find your question. – Christian E. Ramirez Feb 25 '23 at 09:37
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    You made a mistake when applying L'Hospital. There should be only the first term of your sum. The derivative of a constant is $0.$ – Anne Bauval Feb 25 '23 at 09:39
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    Ohhh. Darn. Thanks so much. I just now realized that the second term was a constant value. – Alexander Aseme Feb 25 '23 at 09:40
  • Your answer for the second part is fine. You could improve the proof that $I_0(t)$ is unbounded by providing explicitely a sequence $t_n\to+\infty$ such that $\sin(w_0t_n)=1$ for all $n.$ – Anne Bauval Feb 25 '23 at 09:50
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    Thank you! I wasn't sure how to prove that mathematically at first. – Alexander Aseme Feb 25 '23 at 09:52
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    Note that you did not need to apply first $\sin a\sin b=\dots.$ You could write directly $I(t)=f(t)g(t)$ with (as $w\to w_0$) $f(t)=\frac{2A}{w_{0}+w}\sin\frac{(w_{0}+w)t}{2}\to\frac A{w_0}\sin(w_0t)$ and (by L'Hospital or whatever) $g(t)=\frac1{w_{0}-w}\sin\frac{(w_{0}-w)t}2\to\frac t2.$ – Anne Bauval Feb 25 '23 at 10:05
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    Wow. That's so much easier. I completely forgot about limit laws. It's been so long since I've used them. – Alexander Aseme Feb 25 '23 at 10:19

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