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 $\frac{1}{2}$
 A: You are almost done!
$$\frac{1-\cos \theta}{\theta \sin \theta} = \frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin\theta}=\frac{1-\cos^2\theta}{\theta(1+\cos\theta)\sin \theta}=$$
$$=\frac{\sin\theta}{\theta}\cdot\frac{1}{1+\cos\theta} \to 1\cdot \frac{1}{1+1} = \frac{1}{2}$$
Note (and prove) that indeed
$$\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$$
A: Just to show how to solve with Taylor series: as $\theta \to 0$:
$$\cos(\theta) \approx 1 - \dfrac{\theta^2}{2}$$
$$\sin(\theta) \approx \theta$$
Whence
$$\lim_{\theta \to 0} \frac{1 - \left(1 - \dfrac{\theta^2}{2}\right)}{\theta^2} = \lim_{\theta\to 0} \dfrac{1}{2} = \dfrac{1}{2}$$
A: 
Given that as $\theta$ approaches $0$ then $\dfrac{\sin \theta}{\theta}\approx 0$

should be

Given that as $\theta$ approaches $0$ then $\dfrac{\sin \theta}{\theta}\approx 1$

More precisely, you have
$$
\lim_{\theta \to 0}\frac{\sin \theta}{\theta}=1
$$
Now you can fix your calculation.
A: 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:
$$\lim_{\theta\to0}\frac{2\times(\frac{\theta}2)^2}{\theta\times\theta}=\frac12$$
A: Here it is a slightly different approach from @amirali.
According to the trigonometric identity $\sin(2x) = 2\sin(x)\cos(x)$, we conclude that
\begin{align*}
\lim_{\theta\to0}\frac{1-\cos(\theta)}{\theta\sin(\theta)} & = \lim_{\theta\to 0}\frac{2\sin^{2}(\theta/2)}{2\theta\sin(\theta/2)\cos(\theta/2)}\\\\
& = \lim_{\theta\to 0}\left[\frac{1}{2}\times\frac{\sin(\theta/2)}{\theta/2}\times\frac{1}{\cos(\theta/2)}\right] = \frac{1}{2}
\end{align*}
where we have applied the continuity of $\cos(x)$ at $x = 0$ as well as the following result:
\begin{align*}
\lim_{x\to0}\frac{\sin(x)}{x} = 1
\end{align*}
Hopefully this helps!
