Problem 15 - Olympiad

Using the identity 1 - cos(2x) = 2 sin^2(x): (1 - cos(2x)) / x^2 = 2 sin^2(x) / x^2 = 2 (sin(x)/x)^2. We know lim_(x→0) sin(x)/x = 1. Therefore the limit is 2 · (1)^2 = 2. Alternatively, using L'Hôpital's rule twice: First application: numerator' = 2 sin(2x), denominator' = 2x. Still 0/0. Second application: numerator'' = 4 cos(2x), denominator'' = 2. At x=0: 4(1)/2 = 2. The limit is 2.