Question

If $\lim _{x \rightarrow \infty}\left[\frac{x^3+1}{x^2+1}-(a x+b)\right]=2$, 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

$\lim _{x \rightarrow \infty} \frac{\left[x^3(1-a)-b x^2-a x+(1-b)\right]}{x^2+1}=2$

Limit will exist, if $1-a=0$ and $-b=2$

$\begin{array}{ll}\Rightarrow & a=1 \text { and } & b=-2\end{array}$

So, The correct option will be (C).