Question:

Step 1: Analyze the behavior of $x_n$ as $n \to \infty$. For large values of $n$, the equation $x^n = x^2 + x + 1$ requires $x^n$ to equal a fixed value on the right side. Since $x^n$ grows rapidly for $x > 1$ and decays for $x < 1$, the positive root $x_n$ must approach a value close to 1. Specifically, $x_n \to 1^+$ as $n \to \infty$. Step 2: Express $x_n$ in terms of a small perturbation. Let us write $x_n = 1 + \epsilon_n$ where $\epsilon_n$ is a small positive quantity that approaches 0 as $n \to \infty$. Step 3: Approximate $x_n^n$ using the exponential form. For small $\epsilon_n$, we can use the binomial approximation: $$x_n^n = (1 + \epsilon_n)^n \approx e^{n\epsilon_n}$$ This approximation is valid because $(1 + \epsilon_n)^n \approx e^{n\ln(1+\epsilon_n)} \approx e^{n\epsilon_n}$ when $\epsilon_n$ is small. Step 4: Approximate the right-hand side of the equation. Substituting $x_n = 1 + \epsilon_n$ into $x_n^2 + x_n + 1$: $$x_n^2 + x_n + 1 = (1 + \epsilon_n)^2 + (1 + \epsilon_n) + 1$$ Expanding and keeping only first-order terms in $\epsilon_n$: $$= 1 + 2\epsilon_n + \epsilon_n^2 + 1 + \epsilon_n + 1 \approx 3 + 3\epsilon_n$$ As $\epsilon_n \to 0$, this approaches 3. Step 5: Equate the two sides and solve for $n\epsilon_n$. From the original equation $x_n^n = x_n^2 + x_n + 1$, we have: $$e^{n\epsilon_n} \approx 3 + 3\epsilon_n$$ For small $\epsilon_n$, the term $3\epsilon_n$ is negligible compared to 3, so: $$e^{n\epsilon_n} \approx 3$$ Taking the natural logarithm of both sides: $$n\epsilon_n \approx \ln 3$$ Step 6: Find the limit of $n(x_n - 1)$. Since $\epsilon_n = x_n - 1$, we have: $$\lim_{n \to \infty} n(x_n - 1) = \lim_{n \to \infty} n\epsilon_n = \ln 3$$ Step 7: Calculate the final answer. We need to find $e^{\lim_{n\to\infty} n(x_n - 1)}$: $$e^{\lim_{n\to\infty} n(x_n - 1)} = e^{\ln 3} = 3$$ **Final Answer:** The value of $e^{\left(\lim_{n\to\infty} n(x_n - 1)\right)} = 3$ This corresponds to **Option 3: 3**