calculate the limit $\lim\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$ I'm working on finding the limit for this equation, and would kindly welcome your support to my solution:
$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$$
These are my steps in hopefully deriving the correct result:
$$\frac{1-\cos \theta}{\theta \sin \theta} = \frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin\theta}=\frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin \theta}=\frac{\sin\theta}{\theta}\cdot\frac{1}{1+\cos\theta}$$
Given that as $\theta$ approaches $0$ then $\dfrac{\sin\theta}{\theta}$ $\approx$ $0$ and $\cos\theta \approx 1$. I thought the answer would be $0$ given $\sin{\theta}$ tends to $0$ but the answer is $\frac{1}{2}$. Can someone kindly explain and show this to me? Maybe I went wrong in my reasoning or calculation?
EDIT:
Given that:
$\cos \theta < \frac{\theta}{\sin \theta} < \frac{1}{cos \theta} = \sec \theta$
because $\frac{\theta}{\sin \theta}$ is between two variables approaching 1, then it must also approach 1.
hence, $\frac{\sin \theta}{\theta}$ approaches 1.
So the answer is $\frac{1}{2}$

Note that $1-\cos\theta=2\sin^2\frac{\theta}2$:
$$\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}=\lim_{\theta\to0}\frac{2\sin^2\frac{\theta}2}{\theta \sin\theta}$$
Also we have $\sin u\sim u$ when $u\to0$ so:
$$\lim_{\theta\to0}\frac{2\times(\frac{\theta}2)^2}{\theta\times\theta}=\frac12$$
