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

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

$\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \left(\sqrt{x^2-x+1}-ax\right)\times \frac{\sqrt{x^2-x+1}+ax}{\sqrt{x^2-x+1}+ax}=b$

$\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{\left(x^2-x+1\right)-(ax{)}^2}{\sqrt{x^2-x+1}+ax}=b$

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

$\therefore a=1,-1$

$\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{-x+1}{\sqrt{x^2-x+1}+ax}=b$

$\Rightarrow \underset{x\to \mathrm{\infty }}{lim} \frac{-1+\frac{1}{x}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}+a}=b$

$\Rightarrow \frac{-1}{1+a}=b$

Here, $a\ne -1$

So$a=1$

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

$(a,b)\equiv \left(1,\frac{-1}{2}\right)$