Limits, Continuity & Differentiability
Limits using L'Hôpital Rule
Grade 12
Question:
<p>\(\lim_{x \to 0} \dfrac{e^{x^2} - \cos x}{\sin^2 x}\)</p>
<p>(1) 1</p>
<p>(2) \(\dfrac{3}{2}\)</p>
<p>(3) 2</p>
<p>(4) \(\dfrac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: Use Taylor series expansions around x=0 for e^(x²), cos(x), and sin²(x) to resolve the 0/0 indeterminate form by comparing leading terms.
<p><strong>Step 1:</strong> Verify indeterminate form: As x → 0, numerator → 1 - 1 = 0 and denominator → 0, giving 0/0 form.</p><p><strong>Step 2:</strong> Expand using Taylor series around x = 0:</p><ul><li>e^(x²) = 1 + x² + x⁴/2! + ...</li><li>cos(x) = 1 - x²/2! + x⁴/4! - ...</li><li>sin²(x) = (x - x³/3! + ...)² = x² - x⁴/3 + ...</li></ul><p><strong>Step 3:</strong> Compute numerator: e^(x²) - cos(x) = (1 + x² + x⁴/2 + ...) - (1 - x²/2 + x⁴/24 - ...) = x² + x²/2 + x⁴/2 - x⁴/24 + ... = (3x²/2) + (11x⁴/24) + ...</p><p><strong>Step 4:</strong> Form the limit: (3x²/2 + 11x⁴/24 + ...)/(x² - x⁴/3 + ...) = (3/2 + 11x²/24 + ...)/(1 - x²/3 + ...)</p><p><strong>Step 5:</strong> As x → 0, the limit approaches 3/2.</p><p>∴ Answer: <strong>3/2</strong></p>
Correct Answer: B