<p>\(e^{(x-1) - \frac{1}{2}(x-1)^2 + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots}\) equal to</p>
Step-by-Step Solution
Key Concept: Recognize that the exponent is the Taylor series for ln(1+u) where u = (x-1). The series -u + u²/2 - u³/3 + u⁴/4 - ... = ln(1+u) when |u| < 1. Thus the exponent equals ln(x), making e^(ln x) = x.
<p><strong>Step 1:</strong> Identify the exponent pattern.</p><p>The exponent is: $(x-1) - \frac{1}{2}(x-1)^2 + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \cdots$</p><p><strong>Step 2:</strong> Recognize the Taylor series for $\ln(1+u)$.</p><p>Recall: $\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots$ for $|u| < 1$</p><p><strong>Step 3:</strong> Match with $u = (x-1)$.</p><p>The exponent = $\ln(1+(x-1)) = \ln(x)$ (for $0 < x < 2$)</p><p><strong>Step 4:</strong> Evaluate the exponential.</p><p>$e^{\ln(x)} = x$</p><p>∴ Answer: <strong>x</strong></p>
Correct Answer: C