# Limit of $\frac{1-\cos x}{\sin x}$

I don't understand the rewriting that's being done in this limit:

$$\lim_{x\to0} \frac{1−\cos x}{\sin x} = \lim_{x\to0} \frac{\sin x}{\cos x}$$

Why doesn't this simplify to $\frac{\sin x}{\sin x}$?

• L'Hopitals Rule. – ireallydonknow Nov 28 '14 at 11:43
• and $\sin x/\sin x = 1$ whereas the original limit is $0$ – Ilya Nov 28 '14 at 11:43
• Welcome to Math.SE! Please take a look at math.stackexchange.com/help/notation to see how to use MathJax on this site. – AlexR Nov 28 '14 at 11:47

It doesn't. You can use l'Hospital to get $$\lim_{x\to0} \frac{1-\cos x}{\sin x} = \lim_{x\to0} \frac{\sin x}{\cos x} = \tan 0 = 0$$
rewrite it as $$\frac{1-\cos(x)}{\sin(x)}\frac{1+\cos(x)}{1+\cos(x)}$$
• It's a good idea to find the value of the limit, but it boils down to $\frac{\sin x}{1+\cos x}$ which is something that the OP got – Ilya Nov 28 '14 at 11:51
$$\frac{1-\cos x}{\sin x}=\frac{2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)}=\tan(x/2).$$
Another idea is $$\frac{1-\cos x}{\sin x} = \frac{1-\cos x}{x}\cdot\frac{x}{\sin x}=\frac{1-\cos x}{x}\cdot\frac{1}{\frac{\sin x}{x}}$$