1. **State the problem:** Find the limit as $x \to \pm \infty$ of the expression $$\sqrt{x^2 - 6x - 8} + x - 3.$$\n\n2. **Recall the formula and rules:** When dealing with limits involving square roots of quadratic expressions as $x \to \pm \infty$, it is useful to factor out $x^2$ inside the root to simplify. Also, remember that $$\sqrt{x^2} = |x|,$$ which behaves differently for $x \to +\infty$ and $x \to -\infty$.\n\n3. **Simplify inside the square root:**\n$$\sqrt{x^2 - 6x - 8} = \sqrt{x^2\left(1 - \frac{6}{x} - \frac{8}{x^2}\right)} = |x| \sqrt{1 - \frac{6}{x} - \frac{8}{x^2}}.$$\n\n4. **Consider the limit as $x \to +\infty$: $|x| = x$.**\n$$\lim_{x \to +\infty} \left(\sqrt{x^2 - 6x - 8} + x - 3\right) = \lim_{x \to +\infty} \left(x \sqrt{1 - \frac{6}{x} - \frac{8}{x^2}} + x - 3\right).$$\n\n5. **Use binomial expansion for the square root for large $x$: $$\sqrt{1 + u} \approx 1 + \frac{u}{2}$$ for small $u$. Here, $$u = -\frac{6}{x} - \frac{8}{x^2}.$$**\n$$\sqrt{1 - \frac{6}{x} - \frac{8}{x^2}} \approx 1 - \frac{3}{x} - \frac{4}{x^2}.$$\n\n6. **Multiply by $x$: $$x \sqrt{1 - \frac{6}{x} - \frac{8}{x^2}} \approx x \left(1 - \frac{3}{x} - \frac{4}{x^2}\right) = x - 3 - \frac{4}{x}.$$**\n\n7. **Substitute back into the limit expression:**\n$$\lim_{x \to +\infty} \left(x - 3 - \frac{4}{x} + x - 3\right) = \lim_{x \to +\infty} (2x - 6 - \frac{4}{x}).$$\n\n8. **Evaluate the limit:** As $x \to +\infty$, $2x \to +\infty$, $-6$ is constant, and $-\frac{4}{x} \to 0$. So,\n$$\lim_{x \to +\infty} (2x - 6 - \frac{4}{x}) = +\infty.$$\n\n9. **Now consider the limit as $x \to -\infty$: $|x| = -x$.**\n$$\lim_{x \to -\infty} \left(\sqrt{x^2 - 6x - 8} + x - 3\right) = \lim_{x \to -\infty} \left(-x \sqrt{1 - \frac{6}{x} - \frac{8}{x^2}} + x - 3\right).$$\n\n10. **Approximate the square root as before:**\n$$\sqrt{1 - \frac{6}{x} - \frac{8}{x^2}} \approx 1 - \frac{3}{x} - \frac{4}{x^2}.$$\n\n11. **Multiply by $-x$: $$-x \left(1 - \frac{3}{x} - \frac{4}{x^2}\right) = -x + 3 + \frac{4}{x}.$$**\n\n12. **Substitute back:**\n$$\lim_{x \to -\infty} \left(-x + 3 + \frac{4}{x} + x - 3\right) = \lim_{x \to -\infty} \frac{4}{x} = 0.$$\n\n**Final answers:**\n$$\lim_{x \to +\infty} \left(\sqrt{x^2 - 6x - 8} + x - 3\right) = +\infty,$$\n$$\lim_{x \to -\infty} \left(\sqrt{x^2 - 6x - 8} + x - 3\right) = 0.$$