Question

$\underset{x\to 0}{lim}(\frac{x^2+x+1}{x+1}-ax-b)=4,then$

• 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,

$\underset{x\to 0}{lim}(\frac{x^2+x+1}{x+1}-ax-b)=4$

$\underset{x\to 0}{lim}\left(\frac{x^2+x+1-a{x}^2-ax-bx-b}{x+1}\right)=4$

apply L-Hospital's rule

$\underset{x\to 0}{lim}\left(\frac{2x+1-2ax-a-b}{1}\right)=4$

$\underset{x\to 0}{lim}\left(\frac{x(2-2a)+1-a-b}{1}\right)=4$

$\Rightarrow 2-2a=0$

$\therefore a=1$

and $1-a-b=4$

$b=-3-a$

$\therefore b=-4$