1. Stating the problem: We need to find the limit of the function $$F(x) = x + 1 - \sqrt{x^2 - x - 2}$$ as $$x$$ approaches infinity. 2. Important formula and rules: When dealing with limits involving square roots and polynomials as $$x \to \infty$$, it is useful to factor out the highest power of $$x$$ inside the square root to simplify the expression. 3. Simplify the expression inside the square root: $$\sqrt{x^2 - x - 2} = \sqrt{x^2\left(1 - \frac{1}{x} - \frac{2}{x^2}\right)} = |x| \sqrt{1 - \frac{1}{x} - \frac{2}{x^2}}$$ 4. Since $$x \to \infty$$, $$|x| = x$$, so: $$\sqrt{x^2 - x - 2} = x \sqrt{1 - \frac{1}{x} - \frac{2}{x^2}}$$ 5. Rewrite the original function: $$F(x) = x + 1 - x \sqrt{1 - \frac{1}{x} - \frac{2}{x^2}}$$ 6. Factor out $$x$$: $$F(x) = x \left(1 - \sqrt{1 - \frac{1}{x} - \frac{2}{x^2}}\right) + 1$$ 7. Use the binomial expansion for the square root for large $$x$$: $$\sqrt{1 + u} \approx 1 + \frac{u}{2} - \frac{u^2}{8}$$ where $$u = -\frac{1}{x} - \frac{2}{x^2}$$ 8. Substitute $$u$$: $$\sqrt{1 - \frac{1}{x} - \frac{2}{x^2}} \approx 1 - \frac{1}{2x} - \frac{2}{2x^2} - \frac{\left(-\frac{1}{x} - \frac{2}{x^2}\right)^2}{8}$$ 9. Simplify the main terms: $$\sqrt{1 - \frac{1}{x} - \frac{2}{x^2}} \approx 1 - \frac{1}{2x} - \frac{1}{x^2} + O\left(\frac{1}{x^3}\right)$$ 10. Substitute back into $$F(x)$$: $$F(x) \approx x \left(1 - \left(1 - \frac{1}{2x} - \frac{1}{x^2}\right)\right) + 1 = x \left(\frac{1}{2x} + \frac{1}{x^2}\right) + 1 = \frac{1}{2} + \frac{1}{x} + 1$$ 11. Take the limit as $$x \to \infty$$: $$\lim_{x \to \infty} F(x) = \frac{1}{2} + 0 + 1 = \frac{3}{2}$$ Final answer: $$\boxed{\frac{3}{2}}$$