Solve this limit without using L'Hôpital's rule

Question

I am given the following limit to solve, presumably without knowledge of L'Hôpital's rule:

$$\lim_{x\to0}\left({\frac{x}{1-\cos x}}\right)^2$$

I tried using trigonometric identities (namely Pythagorean) to solve it, but with no luck.

Answer 1

$\displaystyle \lim_{x\to0}\left({\frac{x}{1-\cos x}}\right)^2=$

$\displaystyle = \lim_{x\to0}\left({\frac{x}{2\sin^2 \frac x2}}\right)^2=$

$\displaystyle = \lim_{x\to0}\left({\frac{ \frac{x}{2}}{\sin^2 \frac x2}}\right)^2=$

$\displaystyle = \lim_{x\to0}\left({\frac{ \frac{x}{2}}{\sin \frac x2}} \cdot \frac{1}{\sin \frac x2}\right)^2=$

$\displaystyle =\left(1\cdot \frac{1}{0} \right)^2=$

$=\infty$

Answer 2

$$\frac x{1-\cos x}=\frac x{1-\cos x}\cdot\frac{1+\cos x}{1+\cos x}=\frac{x(1+\cos x)}{\sin^2x}$$