The value of $\lim\limits_{x→0}\left(\frac{1-cosx}{2x^2}\right)$ is :

1 $\frac{1}{2}$ $\frac{1}{4}$ 0

$\frac{1}{4}$

The correct answer is Option (3) → $\frac{1}{4}$ $\lim\limits_{x→0}\frac{1-\cos x}{2x^2}=\lim\limits_{x→0}\frac{2\sin^2\frac{x}{2}}{2x^2}$ $=\lim\limits_{x→0}\frac{\sin^2(\frac{x}{2})}{(\frac{x}{2})^24}=\frac{1}{4}$