trigonometry - 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 $\frac{\sin x}{\sin x}$?

Answer

It doesn't. You can use l'Hospital to get
$$\lim_{x\to0} \frac{1-\cos x}{\sin x} = \lim_{x\to0} \frac{\sin x}{\cos x} = \tan 0 = 0$$