If $$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} - ax} \right) = b$$, then the ordered pair (a, b) is :
A
$$\left( {1,{1 \over 2}} \right)$$
B
$$\left( {1, - {1 \over 2}} \right)$$
C
$$\left( { - 1,{1 \over 2}} \right)$$
D
$$\left( { - 1, - {1 \over 2}} \right)$$
Explanation
$$\mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{x^2} - x + 1} } \right) - ax = b$$ ($$\infty$$ $$-$$ $$\infty$$)
Now, $$\mathop {\lim }\limits_{x \to \infty } {{({x^2} - x + 1 - {a^2}{x^2}}) \over {\sqrt {{x^2} - x + 1} + ax}} = b$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a^2}){x^2} - x + 1} \over {\sqrt {{x^2} - x + 1} + ax}} = b$$
$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{(1 - {a^2}){x^2} - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b$$
$$ \Rightarrow 1 - {a^2} = 0 \Rightarrow a = 1$$
Now, $$\mathop {\lim }\limits_{x \to \infty } {{ - x + 1} \over {x\left( {\sqrt {1 - {1 \over x} + {1 \over {{x^2}}}} + a} \right)}} = b$$
$$ \Rightarrow {{ - 1} \over {1 + a}} = b \Rightarrow b = - {1 \over 2}$$
$$(a,b) = \left( {1, - {1 \over 2}} \right)$$