Problem 5 - Olympiad

First, check continuity at 0: lim_(x→0) x^2 sin(1/x). Since |sin(1/x)| ≤ 1, we have |x^2 sin(1/x)| ≤ x^2. As x → 0, x^2 → 0, so by the Squeeze Theorem, lim_(x→0) x^2 sin(1/x) = 0 = f(0). Thus f is continuous at 0. Now check differentiability: f'(0) = lim_(h→0) [f(h) - f(0)] / h = lim_(h→0) (h^2 sin(1/h)) / h = lim_(h→0) h sin(1/h) = 0 by the Squeeze Theorem (|h sin(1/h)| ≤ |h| → 0). So f'(0) = 0, and f is differentiable at 0. However, f' is not continuous at 0 (since f'(x) = 2x sin(1/x) - cos(1/x) for x ≠ 0, which oscillates). The correct answer is A: f is continuous at x = 0 but not differentiable? Wait, we just showed f'(0) = 0, so it IS differentiable. Let me re-examine. f'(0) = lim h→0 h sin(1/h) = 0. So f is differentiable at 0. So choice A is incorrect. Choice B says f'(0) = 1, which is wrong. Choice C says not continuous, which is wrong. Choice D says the limit does not exist, which is wrong. None of the choices are correct as written. Let me revise the choices or the function. Let me change the function to f(x) = { x sin(1/x) if x ≠ 0; 0 if x = 0 }. Then lim x sin(1/x) = 0, so continuous. f'(0) = lim (h sin(1/h))/h = lim sin(1/h), which does not exist. So f is continuous but not differentiable at 0. Let me revise the question accordingly.