If \(\lim\limits_{x \to \infty}\)\((\sqrt{x^2-x+1} - ax) = b \), then the ordered pair (a, b) is:
(1) \((1, \frac{1}{2})\)
(2) \((1, -\frac{1}{2})\)
(3) \((-1, \frac{1}{2})\)
(4) \((-1, -\frac{1}{2})\)
Correct answer: (2) \((1, -\frac{1}{2})\)
Explanation:
\(\lim\limits_{x \to \infty}\)\((\sqrt{x^2-x+1} - ax) = b \) (∞ - ∞)
⇒ a > 0
Now, \(\lim\limits_{x \to \infty}\)\(\frac{(x^2 - x + 1 - a^2x^2)}{\sqrt{x^2 - x + 1}+ax} \) = b
⇒ \(\lim\limits_{x \to \infty}\)\(\frac{(1 - a^2)x^2 - x + 1}{\sqrt{x^2 - x + 1}+ax} \) = b
⇒ \(\lim\limits_{x \to \infty}\)\(\frac{(1 - a^2)x^2 - x + 1}{x \Big(\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}}+a\Big)} \) = b
⇒ 1 - a2 = 0 ⇒ a = 1
Now, \(\lim\limits_{x \to \infty}\)\(\frac{-x+1}{x \Big(\sqrt{1 - \frac{1}{x} + \frac{1}{x^2}}+a\Big)} \) = b
⇒ \(\frac{-1}{1+a}\) = b = \(-\frac{1}{2}\)
(a, b) = \((1, -\frac{1}{2})\)