1. **State the problem:** Find the limit $$\lim_{x \to -\infty} \frac{x^2 + 6x + 5}{x^5 + 4}$$. 2. **Recall the rule for limits at infinity:** When evaluating limits of rational functions as $x \to \pm \infty$, the behavior is dominated by the highest degree terms in numerator and denominator. 3. **Identify highest degree terms:** Numerator highest degree term is $x^2$, denominator highest degree term is $x^5$. 4. **Divide numerator and denominator by $x^5$ (highest power in denominator):** $$\lim_{x \to -\infty} \frac{\frac{x^2}{x^5} + \frac{6x}{x^5} + \frac{5}{x^5}}{\frac{x^5}{x^5} + \frac{4}{x^5}} = \lim_{x \to -\infty} \frac{x^{-3} + 6x^{-4} + 5x^{-5}}{1 + 4x^{-5}}$$ 5. **Simplify and analyze each term as $x \to -\infty$:** - $x^{-3} = \frac{1}{x^3} \to 0$ - $6x^{-4} = \frac{6}{x^4} \to 0$ - $5x^{-5} = \frac{5}{x^5} \to 0$ - $4x^{-5} = \frac{4}{x^5} \to 0$ 6. **Evaluate the limit:** $$\lim_{x \to -\infty} \frac{0 + 0 + 0}{1 + 0} = \frac{0}{1} = 0$$ **Final answer:** $$\boxed{0}$$