Question

If $\displaystyle\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}-a x\right)=b$, 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)

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