<p>Let \(x_n\) be a positive root of the equation \(x^n = x^2 + x + 1\). Then the value of \(e^{\left(\lim_{n\to\infty} n(x_n - 1)\right)}\) is:</p>
Step-by-Step Solution
Key Concept: As n→∞, x_n → 1⁺ since x^n grows faster than x²+x+1 for x>1. Expand x_n = 1 + ε_n where ε_n is small, substitute into the equation, and use the binomial approximation to find n·ε_n approaches a finite limit.
<p><strong>Step 1:</strong> Observe that x_n satisfies x_n^n = x_n^2 + x_n + 1. As n→∞, if x_n → L > 1, then x_n^n → ∞ but x_n^2 + x_n + 1 → L^2 + L + 1 (finite), which is impossible. If L < 1, then x_n^n → 0 but RHS → L^2 + L + 1 > 0, impossible. Therefore x_n → 1⁺.</p><p><strong>Step 2:</strong> Let x_n = 1 + ε_n where ε_n > 0 is small. Substitute into x_n^n = x_n^2 + x_n + 1:</p><p>(1 + ε_n)^n = (1 + ε_n)^2 + (1 + ε_n) + 1</p><p><strong>Step 3:</strong> Expand LHS using binomial: (1 + ε_n)^n ≈ 1 + nε_n + O(ε_n²) for small ε_n</p><p>Expand RHS: (1 + ε_n)^2 + (1 + ε_n) + 1 = 1 + 2ε_n + ε_n² + 1 + ε_n + 1 = 3 + 3ε_n + ε_n²</p><p><strong>Step 4:</strong> Equate leading terms: 1 + nε_n = 3 + 3ε_n</p><p>Therefore: nε_n - 3ε_n = 2, so ε_n(n - 3) = 2, giving ε_n = 2/(n-3)</p><p><strong>Step 5:</strong> Thus lim_{n→∞} n(x_n - 1) = lim_{n→∞} nε_n = lim_{n→∞} 2n/(n-3) = 2</p><p><strong>Step 6:</strong> Therefore e^(lim_{n→∞} n(x_n-1)) = e² </p><p>∴ Answer: e²</p>
Correct Answer: C