Limits, Continuity & Differentiability
Limits using Taylor series expansion
Grade 12
Question:
<p>If \(\lim_{x \to 0} \dfrac{\cos^2 x - \cos x - e^x \cos x + e^x - \dfrac{x^3}{2}}{x^n} = L\) (where \(L\) is non zero finite), then:</p>
<p>\(L = \dfrac{1}{2}\)</p>
<p>\(n = 3\)</p>
<p>\(L = \dfrac{1}{4}\)</p>
<p>\(n = 4\)</p>
Step-by-Step Solution
Key Concept: Use Taylor series expansion for cos x and e^x around x=0, then combine terms to find the lowest power of x in the numerator that doesn't cancel, determining n such that L is finite and non-zero.
<p><strong>Step 1:</strong> Expand using Taylor series around x = 0:</p><p>cos x = 1 - x²/2 + x⁴/24 - ...</p><p>cos²x = (1 - x²/2 + x⁴/24)² = 1 - x² + x⁴/12 + ...</p><p>e^x = 1 + x + x²/2 + x³/6 + x⁴/24 + ...</p><p>e^x cos x = (1 + x + x²/2 + x³/6 + ...)(1 - x²/2 + x⁴/24 + ...) = 1 + x - x²/2 + x³/6 + x⁴/24 - x⁴/2 + ...</p><p><strong>Step 2:</strong> Compute the numerator:</p><p>cos²x - cos x - e^x cos x + e^x - x³/2</p><p>= (1 - x² + x⁴/12 + ...) - (1 - x²/2 + x⁴/24 - ...) - (1 + x - x²/2 + x³/6 + x⁴/24 + ...) + (1 + x + x²/2 + x³/6 + x⁴/24 + ...) - x³/2</p><p><strong>Step 3:</strong> Collect terms by power:</p><p>Constant: 1 - 1 - 1 + 1 = 0</p><p>Coefficient of x: 0 - 0 - 1 + 1 = 0</p><p>Coefficient of x²: -1 + 1/2 + 1/2 - 1/2 = 0</p><p>Coefficient of x³: 0 - 0 - 1/6 + 1/6 - 1/2 = -1/2</p><p><strong>Step 4:</strong> The first non-zero term is -x³/2, so the numerator behaves as -x³/2 + O(x⁴).</p><p>For L to be finite and non-zero: n = 3</p><p>∴ Answer: C (n = 3, L = -1/2)</p>
Correct Answer: C