Q. If $\lim_{x \rightarrow \infty}{\left\lbrack \frac{x^{3} + 1}{x^{2} + 1} - (ax + b) \right\rbrack = 2}$, then
A) $a = 1$ and $b = 1$
B) $a = 1$ and $b = - 1$
C) $a = 1$ and $b = - 2$
D) $a = 1$ and $b = 2$
Explanation
(c)
Given, $\lim_{x \rightarrow \infty}{\left\lbrack \frac{x^{3} + 1}{x^{2} + 1} - (ax + b) \right\rbrack = 2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \lim_{x \rightarrow \infty}{\left\lbrack \frac{x(1 - a) - b - \frac{a}{x} + \frac{(1 - b)}{x^{2}}}{1 + \frac{1}{x^{2}}} \right\rbrack = 2}$
This limit will exist, if
$1 - a = 0$
and $b = - 2$
$\Rightarrow \ \ \ \ \ \ \ a = 1$
and $b = - 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.