Nihar is ill

The clue is indeed in the problem :) Nihar went to the hospital so we should apply the L'Hopital's Rule . Substituting $x=0$ , the limit takes the form $\dfrac{0}{0}$ , differentiating the numerator and denominator twice we get $\dfrac{\cos(x)}{2}$ . Now we get the required limit by substituting $x=0$ and we get it as $\color{#D61F06}{\boxed{\dfrac{1}{2}}}$ .

$\lim_{x\to 0} \dfrac {1 - \cos x}{x^2}$

$=\lim_{x\to 0} \dfrac {2(1 - \cos x)}{2x^2}$

$=\lim_{x\to 0} \dfrac{2\sin^2(x/2)}{x^2}$

$=\lim_{x\to 0} \dfrac{\sin^2(x/2)}{2\times \dfrac{x^2}{4}}$

$=\dfrac{1}{2} \lim_{x\to 0} \dfrac{\sin^2(x/2)}{\dfrac{x^2}{4}}$

$=\dfrac{1}{2}\times 1\times 1 = \boxed{\dfrac{1}{2}}$

@Mehul Arora , the clue couldn't have been more revealing.