1. **Problem Statement:** Find the values of $a$ and $b$ given that $$\lim_{x \to \infty} \sqrt{x^2 - x + 1 - 2x} = b$$ and $b$ is rational. 2. **Rewrite the expression inside the square root:** $$x^2 - x + 1 - 2x = x^2 - 3x + 1$$ 3. **Factor out $x^2$ inside the square root to simplify the limit:** $$\sqrt{x^2 - 3x + 1} = \sqrt{x^2\left(1 - \frac{3}{x} + \frac{1}{x^2}\right)} = |x| \sqrt{1 - \frac{3}{x} + \frac{1}{x^2}}$$ 4. **As $x \to \infty$, $|x| = x$ since $x$ is positive, so:** $$\sqrt{x^2 - 3x + 1} = x \sqrt{1 - \frac{3}{x} + \frac{1}{x^2}}$$ 5. **Use binomial expansion for the square root for large $x$:** $$\sqrt{1 - \frac{3}{x} + \frac{1}{x^2}} \approx 1 + \frac{a}{x} + \frac{b}{x^2}$$ 6. **Find $a$ by expanding:** $$\left(1 + \frac{a}{x}\right)^2 = 1 + \frac{2a}{x} + \frac{a^2}{x^2} \approx 1 - \frac{3}{x} + \frac{1}{x^2}$$ Matching coefficients: - Coefficient of $\frac{1}{x}$: $2a = -3 \Rightarrow a = -\frac{3}{2}$ - Coefficient of $\frac{1}{x^2}$: $a^2 = 1 \Rightarrow \left(-\frac{3}{2}\right)^2 = \frac{9}{4} \neq 1$, so include $b$ term for accuracy. 7. **Include $b$ term:** $$\sqrt{1 - \frac{3}{x} + \frac{1}{x^2}} \approx 1 + \frac{a}{x} + \frac{b}{x^2}$$ Square both sides: $$1 - \frac{3}{x} + \frac{1}{x^2} \approx 1 + \frac{2a}{x} + \frac{a^2 + 2b}{x^2}$$ Match coefficients: - $\frac{1}{x}$: $2a = -3 \Rightarrow a = -\frac{3}{2}$ - $\frac{1}{x^2}$: $a^2 + 2b = 1 \Rightarrow \left(-\frac{3}{2}\right)^2 + 2b = 1 \Rightarrow \frac{9}{4} + 2b = 1 \Rightarrow 2b = 1 - \frac{9}{4} = -\frac{5}{4} \Rightarrow b = -\frac{5}{8}$ 8. **Rewrite the original limit using this expansion:** $$\lim_{x \to \infty} \sqrt{x^2 - 3x + 1} = \lim_{x \to \infty} x \left(1 + \frac{a}{x} + \frac{b}{x^2}\right) = \lim_{x \to \infty} (x + a + \frac{b}{x}) = \infty$$ This diverges to infinity, so the problem likely involves a different expression or a missing term. 9. **Check the original problem expression carefully:** Given expression is $$\sqrt{x^2 - x + 1 - 2x}$$ which simplifies to $$\sqrt{x^2 - 3x + 1}$$ 10. **If the problem involves $a$ as the coefficient of $x$ inside the root, rewrite as:** $$\sqrt{x^2 + a x + 1 - 2x} = \sqrt{x^2 + (a - 2)x + 1}$$ 11. **Set $a - 2 = -3$ to match the expression inside the root:** $$a - 2 = -3 \Rightarrow a = -1$$ 12. **Now find the limit:** $$\lim_{x \to \infty} \sqrt{x^2 - 3x + 1} - x$$ Rewrite: $$\sqrt{x^2 - 3x + 1} - x = \frac{(x^2 - 3x + 1) - x^2}{\sqrt{x^2 - 3x + 1} + x} = \frac{-3x + 1}{\sqrt{x^2 - 3x + 1} + x}$$ 13. **Divide numerator and denominator by $x$:** $$= \frac{-3 + \frac{1}{x}}{\sqrt{1 - \frac{3}{x} + \frac{1}{x^2}} + 1}$$ 14. **As $x \to \infty$, this becomes:** $$= \frac{-3}{1 + 1} = \frac{-3}{2} = -\frac{3}{2}$$ 15. **Therefore, the limit is:** $$b = -\frac{3}{2}$$ 16. **But the problem states $b$ is rational and gives options for $b = \pm \frac{1}{2}$, so check if the problem meant:** $$\lim_{x \to \infty} \left(\sqrt{x^2 - x + 1 - 2x} - a x\right) = b$$ Try $a = 1$: $$\sqrt{x^2 - 3x + 1} - x = b$$ From step 12-14, $b = -\frac{3}{2}$ which is not in options. Try $a = -1$: $$\sqrt{x^2 - 3x + 1} + x = b$$ As $x \to \infty$, this diverges to infinity. 17. **Try to rationalize the expression with $a = 1$ and subtract $x$:** $$\lim_{x \to \infty} \left(\sqrt{x^2 - 3x + 1} - x\right) = b$$ We found $b = -\frac{3}{2}$, but options have $\pm \frac{1}{2}$. 18. **Check if the original problem has a typo or if $a$ is the coefficient of $x$ inside the root:** Given options suggest $a = \pm 1$ and $b = \pm \frac{1}{2}$. 19. **Try $a = 1$ inside the root:** $$\sqrt{x^2 + x + 1 - 2x} = \sqrt{x^2 - x + 1}$$ 20. **Calculate limit:** $$\lim_{x \to \infty} \sqrt{x^2 - x + 1} - x$$ Rationalize: $$= \frac{x^2 - x + 1 - x^2}{\sqrt{x^2 - x + 1} + x} = \frac{-x + 1}{\sqrt{x^2 - x + 1} + x}$$ Divide numerator and denominator by $x$: $$= \frac{-1 + \frac{1}{x}}{\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}} + 1}$$ As $x \to \infty$: $$= \frac{-1}{1 + 1} = -\frac{1}{2}$$ 21. **So for $a = 1$, $b = -\frac{1}{2}$ which matches option (2).** **Final answer:** $$a = 1, \quad b = -\frac{1}{2}$$