Question

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

• A. $\left(1,\frac{1}{2}\right)$
• B. $\left(1,-\frac{1}{2}\right)$
• C. $\left(-1,\frac{1}{2}\right)$
• D. $\left(-1,-\frac{1}{2}\right)$

Answer 1

Answer: (B)

$\underset{x\to \infty }{\lim }\left(\sqrt{x^2-x+1}\right)-ax=b$ $(\infty -\infty )$
$\Rightarrow a>0$
Now, $\underset{x\to \infty }{\lim }\frac{\left(x^2-x+1-a^2{x}^2\right)}{\sqrt{x^2-x+1}+ax}=b$
$\Rightarrow \underset{x\to \infty }{\lim }\frac{\left(1-a^2\right)x^2-x+1}{\sqrt{x^2-x+1}+ax}=b$
$\Rightarrow \underset{x\to \infty }{\lim }\frac{\left(1-a^2\right)x^2-x+1}{x\left(\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}+a\right)}=b$
$\Rightarrow 1-a^2=0\Rightarrow a=1$
Now, $\underset{x\to \infty }{\lim }\frac{-x+1}{x\left(\sqrt{1-\frac{1}{x}+\frac{1}{x^2}+a}\right)}=b$
$\Rightarrow \frac{-1}{1+a}=b$
$\Rightarrow b=-\frac{1}{2}$
$(a,b)=\left(1,-\frac{1}{2}\right)$