Limits, Continuity & Differentiability
Evaluation of Limits
Grade 12
Question:
<p>Find the value of <em>a</em> and <em>b</em> such that \(\lim_{x \to 0} \dfrac{x(1 + a\cos x) - b\sin x}{x^3} = 1\). Then which one of the following could be true</p>
<p>(a) \(a = -5/2, b = -3/2\)</p>
<p>(b) \(a = -5/2, b = -3/4\)</p>
<p>(c) \(a = -5/4, b = -3/2\)</p>
<p>(d) \(a = -7/2, b = -3/2\)</p>
Step-by-Step Solution
Key Concept: For the limit to exist and equal 1 as x→0, the numerator must have a zero of order exactly 3 at x=0. Use Taylor expansions to match coefficients: expand each term and ensure coefficients of x, x², and x³ terms satisfy the constraint.
<p><strong>Step 1:</strong> Expand using Taylor series:</p><p>cos x = 1 - x²/2 + x⁴/24 + ...</p><p>sin x = x - x³/6 + ...</p><p><strong>Step 2:</strong> Substitute into numerator:</p><p>x(1 + a cos x) - b sin x = x(1 + a(1 - x²/2 + ...)) - b(x - x³/6 + ...)</p><p>= x + ax - ax³/2 + ... - bx + bx³/6 + ...</p><p>= x(1 + a - b) + x³(-a/2 + b/6) + ...</p><p><strong>Step 3:</strong> For limit to exist as x→0, coefficient of x must be zero:</p><p>1 + a - b = 0 → <strong>b = 1 + a</strong></p><p><strong>Step 4:</strong> Apply limit condition:</p><p>lim(x→0) [x³(-a/2 + b/6 + ...)]/x³ = 1</p><p>-a/2 + b/6 = 1</p><p><strong>Step 5:</strong> Substitute b = 1 + a:</p><p>-a/2 + (1 + a)/6 = 1</p><p>-3a + 1 + a = 6</p><p>-2a = 5 → <strong>a = -5/2</strong></p><p><strong>Step 6:</strong> Find b:</p><p>b = 1 + (-5/2) = <strong>-3/2</strong></p><p>∴ Answer: A (or verify which option matches a = -5/2, b = -3/2)</p>
Correct Answer: A