How do I evaluate the limit (1 − cos x)/x² as x approaches 0 for JEE?

It is a 0/0 form so direct substitution fails. The cleanest trick is to use the identity 1 − cos x = 2 sin²(x/2). The expression becomes 2 sin²(x/2) divided by x². Rewrite x² as 4 times (x/2)², so you get [2 sin²(x/2)] / [4 (x/2)²], which is (1/2) times [sin(x/2)/(x/2)]². As x approaches 0, x/2 also approaches 0, and the standard limit sin(t)/t approaches 1, so the bracket squared approaches 1. Therefore the whole limit equals 1/2. Alternatively you could apply L'Hopital's rule twice, but the half-angle identity is faster and avoids differentiation errors under exam pressure. Remember this 1/2 result, it appears constantly.