Question

The values of constants a and b so that $\underset{x\to \infty }{lim}(\frac{x^2+1}{x+1}-ax-b)=\frac{1}{2}$, are

• A. $a=1,b=-\frac{3}{2}$
• B. $a=-1,b=\frac{3}{2}$
• C. $a=0,b=0$
• D. $a=2,b=-1$

Answer 1

The correct option is A $a=1,b=-\frac{3}{2}$

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