Question

Find $a$ and $b$ if :$\underset{x\to \infty }{lim}[\sqrt{x^2-x+1}-ax-b]=0$

• A. $a=1,b=-1$
• B. $a=-1,:b=\frac{1}{2}$
• C. $a=1,b=-\frac{1}{2}$
• D. $a=\frac{1}{2},b=-1$

Answer 1

The correct option is B $a=-1,:b=\frac{1}{2}$

Substitute $x=-t$
$=\underset{t\to \infty }{lim}(\sqrt{t^2+t+1}+at-b)=0$
[For $\infty -\infty$ , a must be negative.]

$=\underset{t\to \infty }{lim}\left[\frac{t^2+t+1-(at-b{)}^2}{\sqrt{t^2+t+1}-(at-b)}\right]=0$

$=\underset{t\to \infty }{lim}\left[\frac{t^2(1-a^2)+t(1+2ab)+1-b^2}{\sqrt{t^2+t+1}-(at-b)}\right]$

For limit coefficient of $t^2$ in numerator should be zero
$1-a^2=0$ & $1+2ab=0$

$\Rightarrow a^2=\pm 1\Rightarrow a=-1\Rightarrow b=\frac{1}{2}$
($a=1$ is rejected)