If $$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) = b$$, then the ordered pair (a, b) is :

$$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} } \right) - ax = b$$ ($\infty$ $-$ $\infty$)<br><br>Now, $$\mathop {\lim }\limits_{x \to \infty } {{({x^2} - x + 1 - {a^2}{x^2}}) \over {\sqrt {{x^2} - x + 1} + ax}} = b$$<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a^2}){x^2} - x + 1} \over {\sqrt {{x^2} - x + 1} + ax}} = b$$<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a^2}){x^2} - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b$$<br><br>$\Rightarrow 1 - {a^2} = 0 \Rightarrow a = 1$<br><br>Now, $$\mathop {\lim }\limits_{x \to \infty } {{ - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b$$<br><br>$\Rightarrow {{ - 1} \over {1 + a}} = b \Rightarrow b = - {1 \over 2}$<br><br>$(a,b) = \left( {1, - {1 \over 2}} \right)$