1. **Stating the problem:** We need to find the limit $$\lim_{x \to \infty} \frac{\sqrt{x^2 + 2}}{x - 8}$$. 2. **Formula and rules:** When dealing with limits at infinity involving square roots and polynomials, it is useful to factor out the highest power of $x$ inside the root and in the denominator to simplify the expression. 3. **Simplify the numerator:** $$\sqrt{x^2 + 2} = \sqrt{x^2\left(1 + \frac{2}{x^2}\right)} = |x| \sqrt{1 + \frac{2}{x^2}}$$ Since $x \to \infty$, $|x| = x$, so: $$\sqrt{x^2 + 2} = x \sqrt{1 + \frac{2}{x^2}}$$ 4. **Rewrite the limit:** $$\lim_{x \to \infty} \frac{x \sqrt{1 + \frac{2}{x^2}}}{x - 8}$$ 5. **Factor $x$ from the denominator:** $$\lim_{x \to \infty} \frac{x \sqrt{1 + \frac{2}{x^2}}}{x\left(1 - \frac{8}{x}\right)}$$ 6. **Cancel $x$ in numerator and denominator:** $$\lim_{x \to \infty} \frac{\cancel{x} \sqrt{1 + \frac{2}{x^2}}}{\cancel{x} \left(1 - \frac{8}{x}\right)} = \lim_{x \to \infty} \frac{\sqrt{1 + \frac{2}{x^2}}}{1 - \frac{8}{x}}$$ 7. **Evaluate the limit by substituting $x \to \infty$:** - $\frac{2}{x^2} \to 0$ - $\frac{8}{x} \to 0$ So: $$\frac{\sqrt{1 + 0}}{1 - 0} = \frac{1}{1} = 1$$ **Final answer:** $$\boxed{1}$$