<p>If constants \(a\), \(b\) and \(c\) so that \(\lim_{x\to 0}\dfrac{axe^x - b\log(1+x) + cxe^{-x}}{x^2\sin x} = 2\) then find \((a+b-c)/8\).</p>
Step-by-Step Solution
Key Concept: Since the denominator x²sin x → 0 as x → 0, the numerator must also → 0, and we need to match coefficients using Taylor series expansions up to sufficient order to make the limit finite and equal to 2.
<p><strong>Step 1: Expand each term using Taylor series around x = 0</strong></p><p>• axe^x = ax(1 + x + x²/2 + x³/6 + ...) = ax + ax² + ax³/2 + ax⁴/6 + ...</p><p>• -b·log(1+x) = -b(x - x²/2 + x³/3 - x⁴/4 + ...)</p><p>• cxe^(-x) = cx(1 - x + x²/2 - x³/6 + ...) = cx - cx² + cx³/2 - cx⁴/6 + ...</p><p><strong>Step 2: Combine numerator terms</strong></p><p>Numerator = (a+c)x + (a-b/2-c)x² + (a/2+b/3+c/2)x³ + (a/6-b/4-c/6)x⁴ + ...</p><p><strong>Step 3: Apply necessary conditions (numerator → 0 as x → 0)</strong></p><p>For the limit to exist and be finite:</p><p>• Coefficient of x: a + c = 0 ⟹ c = -a</p><p>• Coefficient of x²: a - b/2 - c = 0 ⟹ a - b/2 + a = 0 ⟹ b = 4a</p><p><strong>Step 4: Expand denominator and evaluate limit</strong></p><p>x²sin x = x²(x - x³/6 + ...) = x³ - x⁵/6 + ...</p><p>Numerator = (a/2 + b/3 + c/2)x³ + ... = (a/2 + 4a/3 - a/2)x³ + ... = (4a/3)x³ + ...</p><p><strong>Step 5: Calculate the limit</strong></p><p>lim(x→0) = (4a/3)x³/x³ = 4a/3 = 2</p><p>Therefore: a = 3/2, b = 6, c = -3/2</p><p><strong>Step 6: Find (a + b - c)/8</strong></p><p>(3/2 + 6 - (-3/2))/8 = (3/2 + 6 + 3/2)/8 = 9/8</p><p>∴ Answer: <strong>9/8</strong></p>
Correct Answer: 9