Question

If $\underset{x\to \infty }{lim}\left[\frac{x^3+1}{x^2+1}-(ax+b)\right]=2$, then

• A. $a=1$ and $b=1$
• B. $a=1$ and $b=-1$
• C. $a=1$ and $b=-2$
• D. $a=1$ and $b=2$

Answer 1

The correct option is C

$a=1$ and $b=-2$

Explanation for the correct option.

Step 1: Simplify the given equation.

$\begin{array}{rcl}\underset{x\to \infty }{lim}\left[\frac{x^3+1}{x^2+1}-(ax+b)\right]& =& 2\\ \Rightarrow \underset{x\to \infty }{lim}\left[\frac{x^3+1-\left\{(ax+b)\left(x^2+1\right)\right\}}{x^2+1}\right]& =& 2\\ & \Rightarrow & \underset{x\to \infty }{lim}\left[\frac{x^3+1-\left\{a{x}^3+ax+b{x}^2+b\right\}}{x^2+1}\right]=2\\ \Rightarrow \underset{x\to \infty }{lim}\left[\frac{x^3+1-a{x}^3-ax-b{x}^2-b}{x^2+1}\right]& =& 2\\ \Rightarrow \underset{x\to \infty }{lim}\left[\frac{x^3\left(1-a\right)+1-ax-b{x}^2-b}{x^2+1}\right]& =& 2\end{array}$

Step 2:Find the value of $a$.

For the limit to exist, the degree of numerator and denominator should be same, so the coefficient of $x^3$ must be $0$.

$\begin{array}{rcl}& \Rightarrow & 1-a=0\\ \Rightarrow a& =& 1\end{array}$

Step 3:Find the value of $b$.

$\begin{array}{rcl}\underset{x\to \infty }{lim}\left[\frac{1-ax-b{x}^2-b}{x^2+1}\right]& =& 2\\ & \Rightarrow & \underset{x\to \infty }{lim}\left[\frac{{\displaystyle \frac{1}{x^2}}-{\displaystyle \frac{a}{x}}-b-{\displaystyle \frac{b}{x^2}}}{1+{\displaystyle \frac{1}{x^2}}}\right]=2\end{array}$

By applying limit, we get

$\begin{array}{rcl}-b& =& 2\\ & \Rightarrow & b=-2\end{array}$

Therefore, $a=1$ and $b=-2$.

Hence, option C is correct.