Question

If $\displaystyle\lim _{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4$, 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
$\displaystyle\lim _{x \rightarrow \infty} \frac{a_{0} x^{n}+a_{1} x^{n-1}+\ldots +a_{n}}{b_{0} x^{m}+b_{1} x^{m-1}+\ldots +b_{m}}= \begin{cases} 0, & \quad \text{if } n < m\\ \frac{a_0}{b_0} & \quad \text{if } n =m \\ +\infty & \quad \text{if } n > m and a_0 b_0 > 0 \\ - \infty & \quad \text{if } n > m and a_0b_0 <0 \end{cases} $
Description of Situation As to make degree of numerator equal to degree of denominator.
$\displaystyle\lim _{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4$
$=\Rightarrow \displaystyle\lim _{x \rightarrow \infty} \frac{x^{2}+x+1-a x^{2}-a x-b x-b}{x+1}=4$
$\Rightarrow \displaystyle\lim _{x \rightarrow \infty} \frac{x^{2}(1-a)+x(1-a-b)+(1-b)}{x+1}=4$
Here, we make degree of numerator= degree of denominator
$\therefore 1-a=0 \Rightarrow a=1$
and $\displaystyle\lim _{x \rightarrow \infty} \frac{x(1-a-b)+(1-b)}{x+1}=4$
$\Rightarrow 1-a-b=4$
$\Rightarrow b=-4[\because(1-a)=0] .$