Q95 / 155 · WBJEE 2022 · Mathematics

$\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-a x-b\right),(a, b \in R)=0$. Then

  • A
    $a=0, b=1$
  • Correct answer
    B
    $a=1, b=-1$
  • C
    $a=-1, b-1$
  • D
    $a=0, b=0$

Solution

$\lim _{x \rightarrow \infty} \frac{x^2+1-a x(x+1)-b(x+1)}{x+1}=0$ $=\lim _{x \rightarrow \infty} \frac{(1-a) x^2-(a+b) x+1-b}{x+1}=0$ for the limit to exist, $\begin{aligned} 1-\mathrm{a} &=0 \text { and }-(\mathrm{a}+\mathrm{b})=0 \\ \therefore \mathrm{a}=1, \mathrm{~b} &=-1 \end{aligned}$

Hence, the correct answer is option: 'B'.

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