Compute lim_{x->0} (1 - cos x)/x^2.

Compute lim_{x->0} (1 - cos x)/x^2.

• A. 0
• B. 1/2
• C. 1
• D. -1/2

Answer: (B)

This limit is governed by how cosine behaves for very small angles. The standard expansion of cosine near 0 is cos x = 1 - x^2/2 + x^4/24 - ..., so 1 - cos x = x^2/2 - x^4/24 + ..., and dividing by x^2 gives 1/2 - x^2/24 + .... As x approaches 0, the higher-order terms vanish, leaving 1/2. Therefore, the limit is 1/2. You can also see this by L'Hôpital: differentiate top and bottom once to get sin x / (2x), and differentiate again to get cos x / 2, which tends to 1/2 as x -> 0.

This limit is governed by how cosine behaves for very small angles. The standard expansion of cosine near 0 is cos x = 1 - x^2/2 + x^4/24 - ..., so 1 - cos x = x^2/2 - x^4/24 + ..., and dividing by x^2 gives 1/2 - x^2/24 + .... As x approaches 0, the higher-order terms vanish, leaving 1/2. Therefore, the limit is 1/2.

You can also see this by L'Hôpital: differentiate top and bottom once to get sin x / (2x), and differentiate again to get cos x / 2, which tends to 1/2 as x -> 0.