<p>Let \(x_n\) be positive root of the equation \(x^n = x^2 + x + 1\). Then the value of \(e^{\left(\lim\limits_{n\to\infty} n(x_n - 1)\right)}\) is:</p>
Step-by-Step Solution
Key Concept: For large n, the root xₙ approaches 1, so substitute xₙ = 1 + εₙ where εₙ → 0, then expand (1 + εₙ)ⁿ using binomial expansion to find the leading term of εₙ which reveals n·εₙ → ln(3).
<p><strong>Step 1:</strong> Note that x₁ = x² + x + 1 gives x² - x + 1 = 0, which has no real root. For large n, the positive root xₙ → 1 since 1ⁿ = 3 ≠ 1 suggests boundary behavior.</p><p><strong>Step 2:</strong> Let xₙ = 1 + εₙ where εₙ → 0. Then (1 + εₙ)ⁿ = (1 + εₙ)² + (1 + εₙ) + 1.</p><p><strong>Step 3:</strong> Expand left side: (1 + εₙ)ⁿ = 1 + nεₙ + O(ε²ₙ). Expand right side: (1 + εₙ)² + (1 + εₙ) + 1 = 3 + 3εₙ + ε²ₙ.</p><p><strong>Step 4:</strong> Equating: 1 + nεₙ + O(ε²ₙ) = 3 + 3εₙ + ε²ₙ. For large n with εₙ = O(1/n), the nεₙ term dominates the 3εₙ term, giving: nεₙ ≈ 2, so nεₙ → 2 is insufficient.</p><p><strong>Step 5:</strong> More carefully: nεₙ + O(ε²ₙ) = 2 + 3εₙ. As εₙ ~ c/n, we have nεₙ → c. Then c + 0 = 2 + 3(c/n) → c = 2 fails. Recalculate: (1+εₙ)ⁿ ≈ eⁿᵋⁿ for small εₙ, so eⁿᵋⁿ = 3 + 3εₙ + O(ε²ₙ). Thus nεₙ → ln(3).</p><p><strong>Step 6:</strong> Therefore, lim(n→∞) n(xₙ - 1) = ln(3).</p><p><strong>Step 7:</strong> e^(ln(3)) = 3.</p><p>∴ Answer: <strong>3</strong></p>
Correct Answer: B