Question

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

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

Answer 1

The correct option is D $a=1$ and $b=-2$

Given $\underset{x\to \infty }{lim}\frac{x^3+1}{x^2+1}-(ax+b)=2$

$\Rightarrow \underset{x\to \infty }{lim}\left(\frac{{(1-a)x}^3-b{x}^2-ax-b+1}{x^2+1}\right)=2$

For the limit to exist, the coefficient of $x^3$ must be zero because if it is not zero then the limit is infinite

$\Rightarrow a=1$

So the given one will reduce to $\underset{x\to \infty }{lim}\left(\frac{-b{x}^2-ax-b+1}{x^2+1}\right)=2$

If we apply the limit , we get $-b=2$

$\Rightarrow b=-2$

Therefore the correct option is $D$