Question

If ${lim}_{x\to \infty }(\frac{x^2+x+1}{x+1}-ax-b)=4$, then

• A. $a=1,b=-4$
• B. $a=2,b=-3$
• C. $a=2,b=3$
• D. $a=1,b=4$

Answer 1

The correct option is A $a=1,b=-4$

Here,
${lim}_{x\to \infty }\frac{(x^2+x+1)-ax(x+1)-b(x+1)}{x+1}=4$
$\Rightarrow {lim}_{x\to \infty }\frac{x^2(1-a)+x(1-a-b)+1-b}{x+1}=4$
This is possible iff
$1-a=0\Rightarrow a=1$
and $1-a-b=4\Rightarrow b=-4$