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I'm going through the first openstax calculus book and i'm struggling to understand something. Let me process this limit as I think it goes and then ask the question about it.

Evalulate $$\require{cancel}\lim_{\Theta \to 0}\frac{1-\cos \Theta}{\sin \Theta}$$

\begin{align} \lim_{\Theta \to 0}\frac{1-\cos \Theta}{\sin \Theta} & = \lim_{\Theta \to 0}\frac{1 -\cos \Theta}{\sin \Theta}\cdot\frac{1 + \cos \Theta}{1 + \cos \Theta} \\ & = \lim_{\Theta \to 0} \frac{1 - \cos^2 \Theta}{\sin \Theta(1+ \cos \Theta)} \\ & = \lim_{\Theta \to 0} \frac{\sin^2 \Theta}{\sin \Theta(1 + \cos \Theta)} \\ & = \lim_{\Theta \to 0}\frac{\sin \Theta}{\sin \Theta}\cdot\frac{\sin \Theta}{1 + \cos \Theta} \\ & = \frac{0}{0}\cdot \frac{0}{2} \\ & = undef \end{align}

I thought this would have been right, however the book says the answer should be $0$. All I can think of is if it goes like this, picking up on third-last line.

\begin{align} \lim_{\Theta \to 0} \frac{\sin^2 \Theta}{\sin \Theta(1 + \cos \Theta)} & = \lim_{\Theta \to 0}\frac{\cancel{\sin \Theta}}{\cancel{\sin \Theta}}\cdot\frac{\sin \Theta}{1 + \cos \Theta} \\ & = \frac{0}{2} \\ & = 0 \end{align}

But I thought that's not right because for example, $$f(x) = \frac{\cancel{(x+3)}}{\cancel{(x+3)}(x-1)}$$

Still has a discontinuity at $x = -3$

So the cancelling out doesn't remove the discontinuity and I thougt this would be so with the limit above also. Or is there some other way they came to $0$ that I'm not seeing?

Thanks.

Bucephalus
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    related https://math.stackexchange.com/q/2628911 – user Feb 19 '24 at 11:40
  • That's really helpful @user . Thankyou. – Bucephalus Feb 19 '24 at 11:46
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    This is indeed the key point we have $\theta \neq 0$ and so the ratio is equal to $1$. – user Feb 19 '24 at 12:08
  • You are not plugging in $\theta = 0$, you are taking the limit as $\theta$ approaches zero. For any non-zero value of theta, no matter how small, $\frac{\sin{\theta}}{\sin{\theta}}=1$. The same applies to $\frac{x+3}{x+3}$. We are not plugging in $x=-3$ nor are we removing the discontinuity. We are taking the limit. – John Douma Feb 19 '24 at 18:03

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In the numerator, I suggest to write $1-\cos(\theta)$ as $2\sin^{2}(\theta/2)$, while the denominator is $2\sin(\theta/2)\cos(\theta/2)$. Then the limit becomes that of $\tan(\theta/2)$ as $\theta$ tends to $0$ which indeed is $0$.

Jean-Luc Boulnois
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