1. **State the problem:** Find the limit as $x$ approaches infinity of the expression $$\sqrt{x^2 - 5x} - x.$$\n\n2. **Recall the formula and rules:** When dealing with limits involving square roots and large $x$, it is helpful 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$$\lim_{x \to \infty} \left(\sqrt{x^2 - 5x} - x\right) = \lim_{x \to \infty} \frac{(\sqrt{x^2 - 5x} - x)(\sqrt{x^2 - 5x} + x)}{\sqrt{x^2 - 5x} + x}.$$\n\n4. **Simplify the numerator:** Using the difference of squares formula, numerator becomes\n$$x^2 - 5x - x^2 = -5x.$$\nSo the expression is\n$$\lim_{x \to \infty} \frac{-5x}{\sqrt{x^2 - 5x} + x}.$$\n\n5. **Factor $x$ inside the square root:**\n$$\sqrt{x^2 - 5x} = \sqrt{x^2(1 - \frac{5}{x})} = x\sqrt{1 - \frac{5}{x}}.$$\n\n6. **Rewrite the denominator:**\n$$\sqrt{x^2 - 5x} + x = x\sqrt{1 - \frac{5}{x}} + x = x\left(\sqrt{1 - \frac{5}{x}} + 1\right).$$\n\n7. **Substitute back:**\n$$\lim_{x \to \infty} \frac{-5x}{x\left(\sqrt{1 - \frac{5}{x}} + 1\right)} = \lim_{x \to \infty} \frac{-5\cancel{x}}{\cancel{x}\left(\sqrt{1 - \frac{5}{x}} + 1\right)} = \lim_{x \to \infty} \frac{-5}{\sqrt{1 - \frac{5}{x}} + 1}.$$\n\n8. **Evaluate the limit as $x \to \infty$:** Since $\frac{5}{x} \to 0$,\n$$\sqrt{1 - 0} + 1 = 1 + 1 = 2.$$\n\n9. **Final answer:**\n$$\lim_{x \to \infty} \left(\sqrt{x^2 - 5x} - x\right) = \frac{-5}{2} = -\frac{5}{2}.$$