Question

lf $\underset{x\to \infty }{lim}(\frac{x^3+1}{x^2+1}-(ax+b\left)\right)=2$, then

• A. $a=1,b=1$
• B. $a=1,b=2$
• C. $a=1,b=-2$
• D. $a=-1,b=-2$

Answer 1

The correct option is C $a=1,b=-2$

$\underset{x\to \infty }{lim}(\frac{x^3+1}{x^2+1}-(ax+b\left)\right)=2$
$\Rightarrow \underset{x\to \infty }{lim}\left(\frac{x^3+1-(ax+b)(x^2+1)}{x^2+1}\right)=2$
$\Rightarrow \underset{x\to \infty }{lim}\left(\frac{x^3(a-1)-b{x}^2-ax-b)}{x^2+1}\right)=2$
$\Rightarrow \underset{x\to \infty }{lim}\left(\frac{x(a-1)-b-\frac{a}{x}-\frac{b}{x^2})}{1+\frac{1}{x^2}}\right)=2$
For above limit to exist, $a=1$

$\therefore b=-2$