$\lim_{x \to 0}\left(\frac{\sin^2(x)}{1-\cos(x)}\right)$ without L’Hopital’s rule

Question

I'm trying to calculate $\lim_{x \to 0}\left(\frac{\sin^2(x)}{1-\cos(x)}\right)$ without L'Hopital's rule.

The trigonometrical identity $\sin^2(x) = \frac{1-\cos(2x)}{2}$ doesn't seem to lead anywhere. I also attempted to calculate using $\cos^2(x) + \sin^2(x) = 1$ without success.

Any ideas?

Answer 1

Consider the following:

$$\sin^2(x) = 1-\cos^2(x) = (1-\cos(x))(1+\cos(x))$$

You can do the cancellation and BABAM! Evaluating the limit after that is easy :D