Question

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

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

Answer 1

Answer: (B)

PLAN $\left(\frac{\infty }{\infty }\right)$ form
$\underset{x\to \infty }{\lim }\frac{a_0{x}^n+a_1{x}^{n-1}+\dots +a_n}{b_0{x}^m+b_1{x}^{m-1}+\dots +b_m}=\{\begin{array}{ll}0,& \text{if }n<m\\ \frac{a_0}{b_0}& \text{if }n=m\\ +\infty & \text{if }n>mand{a}_0{b}_0>0\\ -\infty & \text{if }n>mand{a}_0{b}_0<0\end{array}$
Description of Situation As to make degree of numerator equal to degree of denominator.
$\underset{x\to \infty }{\lim }\left(\frac{x^2+x+1}{x+1}-ax-b\right)=4$
$=\Rightarrow \underset{x\to \infty }{\lim }\frac{x^2+x+1-a{x}^2-ax-bx-b}{x+1}=4$
$\Rightarrow \underset{x\to \infty }{\lim }\frac{x^2(1-a)+x(1-a-b)+(1-b)}{x+1}=4$
Here, we make degree of numerator= degree of denominator
$\therefore 1-a=0\Rightarrow a=1$
and $\underset{x\to \infty }{\lim }\frac{x(1-a-b)+(1-b)}{x+1}=4$
$\Rightarrow 1-a-b=4$
$\Rightarrow b=-4[\because (1-a)=0].$