Since, $$cos2x = 1 - 2sin^{2}x$$
$$2sin^{2}x = 1 - cos2x$$
Similarly, we can write;
$$2sin^{2}\frac{x}{2} = 1 - cosx$$
Now,Â
$$\lim_{x \rightarrow 0} \frac{1 - \cos x}{x^2} =$$
=Â $$\lim_{x \rightarrow 0} \frac{2sin^{2}\frac{x}{2}}{x^2} =$$
= $$\frac{1}{2}\lim_{x \rightarrow 0} (\frac{sin\frac{x}{2}}{\frac{x}{2}})^{2}$$
= $$\frac{1}{2}$$ * 1Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [$$\lim_{x \rightarrow 0} \frac{sin x}{x} = 1$$]
= $$\frac{1}{2}$$
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