1. **State the problem:** Find the limits of the function $$f(x) = \frac{\sqrt{x+5} - 3}{x - 4}$$ as $$x \to \infty$$ and $$x \to -\infty$$. 2. **Recall the limit rules:** When $$x \to \infty$$ or $$x \to -\infty$$, dominant terms in numerator and denominator determine the limit. Also, rationalizing expressions with square roots helps simplify. 3. **Simplify the numerator by rationalizing:** Multiply numerator and denominator by the conjugate $$\sqrt{x+5} + 3$$: $$f(x) = \frac{(\sqrt{x+5} - 3)(\sqrt{x+5} + 3)}{(x - 4)(\sqrt{x+5} + 3)} = \frac{(x+5) - 9}{(x - 4)(\sqrt{x+5} + 3)} = \frac{x - 4}{(x - 4)(\sqrt{x+5} + 3)}$$ 4. **Cancel common factors:** $$f(x) = \frac{\cancel{x - 4}}{\cancel{x - 4}(\sqrt{x+5} + 3)} = \frac{1}{\sqrt{x+5} + 3}$$ 5. **Evaluate the limit as $$x \to \infty$$:** As $$x \to \infty$$, $$\sqrt{x+5} \to \infty$$, so $$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{1}{\sqrt{x+5} + 3} = 0$$ 6. **Evaluate the limit as $$x \to -\infty$$:** The function is only defined for $$x+5 \geq 0$$ because of the square root, so $$x \geq -5$$. Thus, $$x \to -\infty$$ is outside the domain, so the limit does not exist (DNE). **Final answer:** $$\lim_{x \to \infty} f(x) = 0$$ and $$\lim_{x \to -\infty} f(x) = \text{DNE}$$.