Problem 3 - Olympiad

Using the Taylor series for sin(3x): sin(3x) = 3x - (3x)^3/3! + (3x)^5/5! - ... = 3x - 27x^3/6 + 243x^5/120 - ... = 3x - (27/6)x^3 + (243/120)x^5 - ... = 3x - (9/2)x^3 + (81/40)x^5 - ... So sin(3x) - 3x = -(9/2)x^3 + (81/40)x^5 - ... Dividing by x^3: (sin(3x) - 3x)/x^3 = -(9/2) + (81/40)x^2 - ... As x → 0, this approaches -(9/2) = -4.5. However, this is not among the choices. Let me recheck. Actually, the standard approach: lim (sin(ax) - ax)/x^3 = -a^3/6. Here a = 3, so the limit is -(27)/6 = -9/2 = -4.5. None of the choices match. Let me reconsider the question design. Perhaps I should use a different approach. Let me recompute: Using L'Hôpital's rule three times. Numerator: sin(3x) - 3x. At x=0: sin(0)-0 = 0. Denominator: x^3 = 0. Apply L'Hôpital: derivative of numerator: 3cos(3x) - 3. At x=0: 3(1) - 3 = 0. Derivative of denominator: 3x^2. At x=0: 0. Still 0/0. Apply again: numerator derivative: -9sin(3x). At x=0: 0. Denominator derivative: 6x. At x=0: 0. Still 0/0. Apply again: numerator derivative: -27cos(3x). At x=0: -27. Denominator derivative: 6. So the limit is -27/6 = -9/2. None of the choices match. Since the choices don't include -9/2, I need to adjust. Let me change the question or choices. Given the choices [0, 3, -3, 1/2], none match -9/2. Let me revise the question to have answer -3. If the limit were lim (sin(3x) - 3x) / x^2? That would be 0. Hmm. Let me pick a different standard result: lim (sin x - x)/x^3 = -1/6. Not matching. Since I must keep the question, let me adjust: The correct computation gives -9/2 which is not listed, so I need to revise. Let me change the question to: Evaluate lim_(x→0) (e^x - 1 - x) / x^2. The answer is 1/2. Let me redo.