IMO Practice Test — Limits and Derivatives

Explanation: f(x) is x² for x≥0 and −x² for x<0. f'(x) is 2x and −2x respectively. At x=0, both left and right derivatives evaluate to 0. Thus, f'(0) = 0.

Explanation: Factor x⁴−16=(x−2)(x+2)(x²+4). Substituting x=2 gives 4×8=32.

Explanation: This is the definition of the derivative of sin x at x = a, which is cos a. Alternatively, use sum-to-product formulas: 2 cos((x+a)/2) sin((x-a)/2) / (x-a).

Explanation: 1 − cos x = 2 sin²(x/2). The denominator is √(2 sin²(x/2)) = √2 |sin(x/2)|. As x approaches 0 from the right, the limit is √2. From the left, it is −√2. Since LHL ≠ RHL, the limit does not exist.

Explanation: Since −1 ≤ sin(1/x) ≤ 1, we have −|x| ≤ x sin(1/x) ≤ |x|. By the Squeeze Theorem, as x → 0, both bounds go to 0, so the limit is 0.

Explanation: Divide top and bottom inside bracket by 3x: [ (1 + 2/3x) / (1 - 1/3x) ]^(3x). This is [ (1 + 2/3x)^(3x/2) ]² / [ (1 - 1/3x)^(-3x) ]^(-1). Both base limits approach e. We get e² / e^(-1) = e³.