$$\lim_{x \rightarrow 0} \frac{1 - \cos x}{x^2} =$$

Since, $$cos2x = 1 - 2sin^{2}x$$

$$2sin^{2}x = 1 - cos2x$$

Similarly, we can write;

$$2sin^{2}\frac{x}{2} = 1 - cosx$$

Now, 

$$\lim_{x \rightarrow 0} \frac{1 - \cos x}{x^2} =$$

= $$\lim_{x \rightarrow 0} \frac{2sin^{2}\frac{x}{2}}{x^2} =$$

= $$\frac{1}{2}\lim_{x \rightarrow 0} (\frac{sin\frac{x}{2}}{\frac{x}{2}})^{2}$$

= $$\frac{1}{2}$$ * 1                                             [$$\lim_{x \rightarrow 0} \frac{sin x}{x} = 1$$]

= $$\frac{1}{2}$$