Limits, Continuity & Differentiability
Limits using expansions
Grade 12
Question:
<p>If \(\displaystyle\lim_{x \to 0}\left(\dfrac{\sin 3x}{x^3} + \dfrac{a}{x^2} + b\right) = 0\), then the value of \((a + b)\) equals:</p>
<p>(a) 0</p>
<p>(b) \(\dfrac{1}{2}\)</p>
<p>(c) \(\dfrac{3}{2}\)</p>
<p>(d) 3</p>
Step-by-Step Solution
Key Concept: For the limit to exist and equal zero as x→0, the coefficients of negative powers of x in the Laurent expansion must sum to zero. Expand sin(3x) using Taylor series and match powers of x to find a and b.
<p><strong>Step 1:</strong> Expand sin(3x) using Taylor series:</p><p>sin(3x) = 3x - (3x)³/3! + (3x)⁵/5! - ... = 3x - 9x³/2 + 81x⁵/40 - ...</p><p><strong>Step 2:</strong> Divide by x³:</p><p>sin(3x)/x³ = 3/x² - 9/2 + 81x²/40 - ...</p><p><strong>Step 3:</strong> Rewrite the given limit:</p><p>lim(x→0)[3/x² + a/x² - 9/2 + b + 81x²/40 - ...] = 0</p><p><strong>Step 4:</strong> For the limit to exist and equal zero, coefficients of negative powers must vanish:</p><p>Coefficient of 1/x²: 3 + a = 0 ⟹ a = -3</p><p><strong>Step 5:</strong> The constant term must also be zero:</p><p>-9/2 + b = 0 ⟹ b = 9/2</p><p><strong>Step 6:</strong> Calculate a + b:</p><p>a + b = -3 + 9/2 = -6/2 + 9/2 = 3/2</p><p>∴ Answer: C (which equals 3/2)</p>
Correct Answer: C