1. **State the problem:** We want to find the limit $$\lim_{x \to -\infty} \left( \sqrt{x^2 - 5x + 1} - 2x \right).$$ 2. **Recall the formula and approach:** When dealing with limits involving square roots and infinity, a common technique is to rationalize the expression to simplify it. 3. **Rationalize the expression:** Multiply and divide by the conjugate: $$\lim_{x \to -\infty} \left( \sqrt{x^2 - 5x + 1} - 2x \right) \times \frac{\sqrt{x^2 - 5x + 1} + 2x}{\sqrt{x^2 - 5x + 1} + 2x} = \lim_{x \to -\infty} \frac{(x^2 - 5x + 1) - 4x^2}{\sqrt{x^2 - 5x + 1} + 2x}.$$ 4. **Simplify numerator:** $$x^2 - 5x + 1 - 4x^2 = -3x^2 - 5x + 1.$$ 5. **Rewrite the limit:** $$\lim_{x \to -\infty} \frac{-3x^2 - 5x + 1}{\sqrt{x^2 - 5x + 1} + 2x}.$$ 6. **Factor out $x^2$ inside the square root:** $$\sqrt{x^2 - 5x + 1} = \sqrt{x^2\left(1 - \frac{5}{x} + \frac{1}{x^2}\right)} = |x| \sqrt{1 - \frac{5}{x} + \frac{1}{x^2}}.$$ 7. **Since $x \to -\infty$, $|x| = -x$. So:** $$\sqrt{x^2 - 5x + 1} = -x \sqrt{1 - \frac{5}{x} + \frac{1}{x^2}}.$$ 8. **Rewrite denominator:** $$\sqrt{x^2 - 5x + 1} + 2x = -x \sqrt{1 - \frac{5}{x} + \frac{1}{x^2}} + 2x = x \left(- \sqrt{1 - \frac{5}{x} + \frac{1}{x^2}} + 2 \right).$$ 9. **Rewrite the entire expression:** $$\lim_{x \to -\infty} \frac{-3x^2 - 5x + 1}{x \left(- \sqrt{1 - \frac{5}{x} + \frac{1}{x^2}} + 2 \right)} = \lim_{x \to -\infty} \frac{x^2(-3 - \frac{5}{x} + \frac{1}{x^2})}{x \left(- \sqrt{1 - \frac{5}{x} + \frac{1}{x^2}} + 2 \right)} = \lim_{x \to -\infty} \frac{x(-3 - \frac{5}{x} + \frac{1}{x^2})}{- \sqrt{1 - \frac{5}{x} + \frac{1}{x^2}} + 2}.$$ 10. **As $x \to -\infty$, terms with $\frac{1}{x}$ and $\frac{1}{x^2}$ go to zero:** $$\lim_{x \to -\infty} \frac{x(-3 + 0 + 0)}{- \sqrt{1 + 0 + 0} + 2} = \lim_{x \to -\infty} \frac{-3x}{-1 + 2} = \lim_{x \to -\infty} \frac{-3x}{1} = \lim_{x \to -\infty} -3x.$$ 11. **Since $x \to -\infty$, $-3x \to +\infty$.** **Final answer:** $$\lim_{x \to -\infty} \left( \sqrt{x^2 - 5x + 1} - 2x \right) = +\infty.$$