Question Detail

To evaluate limx→0 (x / √(1-cos x)), we can use the identity 1 - cos x = 2sin²(x/2):limx→0 (x / √(2sin²(x/2))) = limx→0 (x / (√2 |sin(x/2)|))Now, consider the one-sided limits:For x → 0+, sin(x/2) > 0, so |sin(x/2)| = sin(x/2). The limit becomes:limx→0+ (x / (√2 sin(x/2))) = limx→0+ (1 / √2) * (x / sin(x/2))Using limu→0 (u / sin u) = 1, let u = x/2. Then x = 2u:limu→0+ (1 / √2) * (2u / sin u) = (1 / √2) * 2 * 1 = 2/√2 = √2For x → 0-, sin(x/2) < 0, so |sin(x/2)| = -sin(x/2). The limit becomes:limx→0- (x / (√2 (-sin(x/2)))) = limx→0- (-1 / √2) * (x / sin(x/2))limu→0- (-1 / √2) * (2u / sin u) = (-1 / √2) * 2 * 1 = -2/√2 = -√2Since the left-hand limit ( -√2 ) and the right-hand limit ( √2 ) are not equal, the limit does not exist. In multiple-choice questions, '∞' (infinity) is often used to indicate that the limit diverges or does not exist.