1. **State the problem:** Find the limit $$\lim_{x \to -\infty} \sqrt{4x^2 - 6} + \sqrt{x^2 + 1}$$. 2. **Recall the rule:** For limits involving square roots of quadratic expressions as $x \to \pm \infty$, factor out $x^2$ inside the root to simplify. 3. **Rewrite each square root:** $$\sqrt{4x^2 - 6} = \sqrt{x^2(4 - \frac{6}{x^2})} = |x| \sqrt{4 - \frac{6}{x^2}}$$ $$\sqrt{x^2 + 1} = \sqrt{x^2(1 + \frac{1}{x^2})} = |x| \sqrt{1 + \frac{1}{x^2}}$$ 4. **Since $x \to -\infty$, $|x| = -x$. Substitute:** $$\sqrt{4x^2 - 6} + \sqrt{x^2 + 1} = -x \sqrt{4 - \frac{6}{x^2}} - x \sqrt{1 + \frac{1}{x^2}}$$ 5. **Factor out $-x$:** $$= -x \left( \sqrt{4 - \frac{6}{x^2}} + \sqrt{1 + \frac{1}{x^2}} \right)$$ 6. **Evaluate the limit inside the parentheses as $x \to -\infty$:** $$\lim_{x \to -\infty} \sqrt{4 - \frac{6}{x^2}} = \sqrt{4 - 0} = 2$$ $$\lim_{x \to -\infty} \sqrt{1 + \frac{1}{x^2}} = \sqrt{1 + 0} = 1$$ 7. **So the expression behaves like:** $$-x (2 + 1) = -x \cdot 3$$ 8. **Since $x \to -\infty$, $-x \to +\infty$, so:** $$\lim_{x \to -\infty} \sqrt{4x^2 - 6} + \sqrt{x^2 + 1} = +\infty$$ **Final answer:** The limit diverges to infinity.