The value of limlimits_x→0≤ft((1-cosx)/(2x^2)) is :

The correct answer is Option (3) → $\frac{1}{4}$

$\lim\limits_{x→0}\frac{1-\cos x}{2x^2}=\lim\limits_{x→0}\frac{2\sin^2\frac{x}{2}}{2x^2}$

$=\lim\limits_{x→0}\frac{\sin^2(\frac{x}{2})}{(\frac{x}{2})^24}=\frac{1}{4}$