<p>\(\lim_{x \to 0} \dfrac{\sin(\pi\cos^2 x)}{x^2}\)</p>
Step-by-Step Solution
Key Concept: Use the substitution that cos²x → 1 as x → 0, so sin(π cos²x) → sin(π) = 0, creating a 0/0 form. Then apply Taylor expansion: cos²x = 1 - x² + O(x⁴), making π cos²x = π - πx² + O(x⁴), and use sin(π - u) ≈ u for small u.
<p><strong>Step 1:</strong> Recognize the indeterminate form. As x → 0: cos²x → 1, so sin(π cos²x) → sin(π) = 0, giving 0/0.</p><p><strong>Step 2:</strong> Use Taylor expansion for cos²x: cos²x = 1 - x² + O(x⁴)</p><p><strong>Step 3:</strong> Substitute: π cos²x = π(1 - x² + O(x⁴)) = π - πx² + O(x⁴)</p><p><strong>Step 4:</strong> Apply sin(π - u) = sin(u) for small u: sin(π - πx²) = sin(πx²)</p><p><strong>Step 5:</strong> Use sin(πx²) ≈ πx² for small πx²:</p><p>$$\lim_{x \to 0} \frac{\sin(\pi\cos^2 x)}{x^2} = \lim_{x \to 0} \frac{\sin(\pi x^2)}{x^2} = \lim_{x \to 0} \frac{\pi x^2}{x^2} = \pi$$</p><p><strong>∴ Answer: B (π)</strong></p>
Correct Answer: B