If _ x → ∞ ( x ^ 2 - x + 1 - ax) = b , then the ordered pair ( a, b...

Given that:$\lim_{x \to \infty}\left(\sqrt{{x}^{2} - x + 1} - ax\right) = b$

$⇒ \lim_{x \to \infty}\left(\sqrt{{x}^{2} - x + 1} - ax\right) \times \dfrac{\sqrt{{x}^{2} - x + 1} + ax}{\sqrt{{x}^{2} - x + 1} + ax} = b$

$⇒ \lim_{x \to \infty}\dfrac{\left({x}^{2} - x + 1\right) - ( ax{)}^{2}}{\sqrt{{x}^{2} - x + 1} + ax} = b$

$⇒ \text{Limit exist only If}{a}^{2} = 1$, because ${x}^{2}$ must not be there in the numerator

$∴ a = 1, - 1$

$⇒ \lim_{x \to \infty}\dfrac{- x + 1}{\sqrt{{x}^{2} - x + 1} + ax} = b$

$⇒ \lim_{x \to \infty}\dfrac{- 1 + \dfrac{1}{x}}{\sqrt{1 - \dfrac{1}{x} + \dfrac{1}{{x}^{2}}} + a} = b$

$⇒ \dfrac{- 1}{1 + a} = b$

Here, $a \ne - 1$

So$a = 1$

$b = \dfrac{- 1}{2}$

$( a, b ) ≡ \left(1, \dfrac{- 1}{2}\right)$