Q58 / 120 · VITEEE 2011 · Mathematics
The values of constants $\mathrm{a}$ and $\mathrm{b}$, so that $\lim _{x \rightarrow \infty}\left(\frac{x^{2}+1}{x+1}-a x-b\right)=0$ are
- A$a=0, b=0$
- Correct answerB$a=1, b=-1$
- C$a=-1, b=1$
- D$a=2, b=-1$
Solution
$$
\begin{array}{l}
\lim _{x \rightarrow \infty}\left(\frac{x^{2}+1}{x+1}-a x-b\right)=0 \\
\Rightarrow \lim _{x \rightarrow \infty}\left[\frac{x^{2}(1-a)-(a+b) x-b+1}{x+1}\right]=0 \\
\Rightarrow 1-a=0 \text { and } a+b=0 \Rightarrow a=1, b=-1
\end{array}
$$
Hence, the correct answer is option: 'B'.