Question

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

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

Answer 1

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

Given: $\underset{x\to \infty }{lim}(\frac{x^2+x+1}{x+1}-ax-b)=4$

$\Rightarrow \underset{x\to \infty }{lim}\left(\frac{x^2+x+1-a{x}^2-ax-bx-b}{x+1}\right)=4$

$\Rightarrow \underset{x\to \infty }{lim}\left(\frac{x^2(1-a)+x(1-a-b)+1-b}{x+1}\right)=4$

This is possible only when

$1-a=0\Rightarrow a=1$

and $1-a-b=4\Rightarrow 1-1-b=4\Rightarrow b=-4$

Therefore, $a=1$ and $b=-4$.