1. **State the problem:** Find the limit as $x \to \infty$ of the expression $$\sqrt{x^2 - 5x} - x.$$\n\n2. **Recall the formula and approach:** When dealing with limits involving square roots and large $x$, a common technique is to rationalize the expression to simplify it.\n\n3. **Rationalize the expression:** Multiply numerator and denominator by the conjugate $$\sqrt{x^2 - 5x} + x$$ to get\n$$\left(\sqrt{x^2 - 5x} - x\right) \cdot \frac{\sqrt{x^2 - 5x} + x}{\sqrt{x^2 - 5x} + x} = \frac{(\sqrt{x^2 - 5x})^2 - x^2}{\sqrt{x^2 - 5x} + x} = \frac{x^2 - 5x - x^2}{\sqrt{x^2 - 5x} + x} = \frac{-5x}{\sqrt{x^2 - 5x} + x}.$$\n\n4. **Simplify the denominator:** Factor $x^2$ inside the square root:\n$$\sqrt{x^2 - 5x} = \sqrt{x^2\left(1 - \frac{5}{x}\right)} = x\sqrt{1 - \frac{5}{x}}.$$\n\n5. **Rewrite the expression:**\n$$\frac{-5x}{x\sqrt{1 - \frac{5}{x}} + x} = \frac{-5x}{x\left(\sqrt{1 - \frac{5}{x}} + 1\right)}.$$\n\n6. **Cancel $x$ in numerator and denominator:**\n$$\frac{-5\cancel{x}}{\cancel{x}\left(\sqrt{1 - \frac{5}{x}} + 1\right)} = \frac{-5}{\sqrt{1 - \frac{5}{x}} + 1}.$$\n\n7. **Evaluate the limit as $x \to \infty$:** Since $\frac{5}{x} \to 0$, we have\n$$\lim_{x \to \infty} \frac{-5}{\sqrt{1 - 0} + 1} = \frac{-5}{1 + 1} = \frac{-5}{2} = -\frac{5}{2}.$$\n\n**Final answer:** $$\boxed{-\frac{5}{2}}.$$