<p>\(\lim_{x \to 0} \dfrac{e^{x^2} - \cos x}{\sin^2 x}\) is equal to</p>
Step-by-Step Solution
Key Concept: Use Taylor series expansion around x = 0 for e^(x²), cos x, and sin x to compare the leading terms in numerator and denominator. Both e^(x²) and cos x expand with 1 as the leading constant term, so their difference gives a meaningful result.
<p><strong>Step 1:</strong> Expand using Taylor series around x = 0:</p><p>e^(x²) = 1 + x² + x⁴/2! + ...</p><p>cos x = 1 - x²/2! + x⁴/24 - ...</p><p>sin x = x - x³/6 + ..., so sin²x = x² - x⁴/3 + ...</p><p><strong>Step 2:</strong> Find the numerator:</p><p>e^(x²) - cos x = (1 + x² + x⁴/2 + ...) - (1 - x²/2 + x⁴/24 - ...)</p><p>= x² + x²/2 + x⁴(1/2 - 1/24) + ...</p><p>= x²(3/2 + x²·11/24 + ...)</p><p><strong>Step 3:</strong> Divide by sin²x = x²(1 - x²/3 + ...):</p><p>lim(x→0) [x²(3/2 + ...)] / [x²(1 - x²/3 + ...)] = 3/2 ÷ 1 = 3/2</p><p>∴ Answer: B</p>
Correct Answer: B