<p>Find \(a\), \(b\) and \(c\) such that \[\lim_{x \to 0} \frac{ax e^x - b\log(1+x) + cx e^{-x}}{x^2 \sin x} = 2\]</p>
Step-by-Step Solution
Key Concept: Use Taylor series expansions for e^x, log(1+x), and e^{-x} around x=0, then match coefficients of powers of x in numerator and denominator to extract conditions and evaluate the limit.
<p><strong>Step 1: Expand all functions using Taylor series around x = 0</strong></p><p>• e^x = 1 + x + x²/2 + x³/6 + ...</p><p>• log(1+x) = x - x²/2 + x³/3 - ...</p><p>• e^{-x} = 1 - x + x²/2 - x³/6 + ...</p><p>• sin(x) = x - x³/6 + ...</p><p><strong>Step 2: Expand the numerator</strong></p><p>axe^x = ax(1 + x + x²/2 + x³/6 + ...) = ax + ax² + ax³/2 + ...</p><p>-b·log(1+x) = -bx + bx²/2 - bx³/3 + ...</p><p>cxe^{-x} = cx(1 - x + x²/2 - x³/6 + ...) = cx - cx² + cx³/2 - ...</p><p>Numerator = (a - b + c)x + (a + b/2 - c)x² + (a/2 - b/3 + c/2)x³ + ...</p><p><strong>Step 3: Expand the denominator</strong></p><p>x²sin(x) = x²(x - x³/6 + ...) = x³ - x⁵/6 + ...</p><p><strong>Step 4: Apply the limit condition</strong></p><p>For the limit to exist and equal 2, the coefficient of x and x² in numerator must be zero:</p><p>• Coefficient of x: a - b + c = 0</p><p>• Coefficient of x²: a + b/2 - c = 0 ⟹ 2a + b - 2c = 0</p><p><strong>Step 5: Verify with the limit value</strong></p><p>With these conditions, numerator = (a/2 - b/3 + c/2)x³ + ... and limit = (a/2 - b/3 + c/2)/1 = 2</p><p>∴ <strong>Conditions:</strong> a - b + c = 0 and 2a + b - 2c = 0</p>
Correct Answer: a - b + c = 0, 2a + b - 2c = 0