1. **State the problem:** Find the limit $$\lim_{x \to \infty} \frac{\sqrt{x} - 3 - \sqrt{5x} + 1}{\sqrt{5x} - 1 - \sqrt{x} + 3}$$. 2. **Rewrite the expression:** Group terms to see dominant behavior: $$\frac{(\sqrt{x} - \sqrt{5x}) + (-3 + 1)}{(\sqrt{5x} - \sqrt{x}) + (-1 + 3)} = \frac{\sqrt{x}(1 - \sqrt{5}) - 2}{\sqrt{x}(\sqrt{5} - 1) + 2}$$. 3. **Divide numerator and denominator by $\sqrt{x}$ to analyze limit as $x \to \infty$:** $$\frac{\sqrt{x}(1 - \sqrt{5}) - 2}{\sqrt{x}(\sqrt{5} - 1) + 2} = \frac{(1 - \sqrt{5}) - \frac{2}{\sqrt{x}}}{(\sqrt{5} - 1) + \frac{2}{\sqrt{x}}}$$. 4. **Evaluate the limit as $x \to \infty$:** Terms with $\frac{2}{\sqrt{x}}$ go to zero, so $$\lim_{x \to \infty} \frac{(1 - \sqrt{5}) - 0}{(\sqrt{5} - 1) + 0} = \frac{1 - \sqrt{5}}{\sqrt{5} - 1}$$. 5. **Simplify the fraction:** Multiply numerator and denominator by $-1$ to get $$\frac{1 - \sqrt{5}}{\sqrt{5} - 1} = \frac{-(\sqrt{5} - 1)}{\sqrt{5} - 1} = -1$$. **Final answer:** $$\boxed{-1}$$. This corresponds to option C.