<p>The value of \(e^{(x-1) - \frac{1}{2}(x-1)^2 + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots}\) is:</p>
Step-by-Step Solution
Key Concept: Recognize the series in the exponent as the Taylor expansion of ln(1+u) where u = (x-1). The series ∑(-1)^(n+1)·u^n/n = ln(1+u) for |u| < 1.
<p><strong>Step 1:</strong> Identify the series in the exponent as the Taylor expansion of ln(1+u).</p><p>The series is: (x-1) - (x-1)²/2 + (x-1)³/3 - (x-1)⁴/4 + ⋯</p><p>This matches the form: u - u²/2 + u³/3 - u⁴/4 + ⋯ = ln(1+u) where u = (x-1)</p><p><strong>Step 2:</strong> Apply the logarithm series formula.</p><p>∑(n=1 to ∞) [(-1)^(n+1) · u^n / n] = ln(1+u)</p><p>Therefore: (x-1) - (x-1)²/2 + (x-1)³/3 - ⋯ = ln(1 + (x-1)) = ln(x)</p><p><strong>Step 3:</strong> Evaluate the exponential.</p><p>e^(ln(x)) = x</p><p>∴ <strong>Answer: C (which equals x)</strong></p>
Correct Answer: C