1. **State the problem:** Find all horizontal asymptotes of the function $$f(x) = \frac{\sqrt{x + 4}}{2x - 3}$$. 2. **Recall the definition of horizontal asymptotes:** Horizontal asymptotes describe the behavior of a function as $$x \to \infty$$ or $$x \to -\infty$$. We find them by evaluating $$\lim_{x \to \infty} f(x)$$ and $$\lim_{x \to -\infty} f(x)$$. 3. **Evaluate the limit as $$x \to \infty$$:** $$\lim_{x \to \infty} \frac{\sqrt{x + 4}}{2x - 3} = \lim_{x \to \infty} \frac{\sqrt{x}\sqrt{1 + \frac{4}{x}}}{2x - 3}$$ Since $$\sqrt{1 + \frac{4}{x}} \to 1$$ as $$x \to \infty$$, this simplifies to: $$\lim_{x \to \infty} \frac{\sqrt{x}}{2x - 3} = \lim_{x \to \infty} \frac{\sqrt{x}}{2x}$$ (ignoring the -3 as it becomes negligible) Rewrite $$\sqrt{x} = x^{1/2}$$, so: $$\lim_{x \to \infty} \frac{x^{1/2}}{2x} = \lim_{x \to \infty} \frac{1}{2} x^{1/2 - 1} = \lim_{x \to \infty} \frac{1}{2} x^{-1/2}$$ Since $$x^{-1/2} = \frac{1}{\sqrt{x}} \to 0$$ as $$x \to \infty$$, the limit is: $$0$$ 4. **Evaluate the limit as $$x \to -\infty$$:** The domain of $$f(x)$$ requires $$x + 4 \geq 0$$, so $$x \geq -4$$. Therefore, $$f(x)$$ is not defined for $$x < -4$$, and the limit as $$x \to -\infty$$ does not exist. 5. **Conclusion:** The only horizontal asymptote is: $$y = 0$$ This means the graph approaches the x-axis as $$x$$ becomes very large.