The values of 'a' and , b So that Limx→∞⁡x2+1x+1−ax−b=0 are

Limy→0+⁡1+y2−(a+by)(1+y)y(1+y)=0⇒a=1 and Limy→0+⁡y2(1−b)−(a+b)yy′−1+y)1+0=0⇒Limy→0+⁡y(1−b)−(a+b)1+y=a+b=0⇒b=−1

The values of 'a' and , b So that Limx→∞⁡x2+1x+1−ax−b=0 are

Limy→0+⁡1+y2−(a+by)(1+y)y(1+y)=0⇒a=1 and Limy→0+⁡y2(1−b)−(a+b)yy′−1+y)1+0=0⇒Limy→0+⁡y(1−b)−(a+b)1+y=a+b=0⇒b=−1

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The values of 'a' and , b So that ${Lim}_{x \to \infty} (\frac{x^{2} + 1}{x + 1} - a x - b) = 0$ are

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1,-1

-1,1

1,1

-1,-1

answer is A.

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Detailed Solution

${Lim}_{y \to 0^{+}} \frac{(1 + y^{2}) - (a + b y) (1 + y)}{y (1 + y)} = 0$
$\begin{array}{l} \Rightarrow a = 1 and \\ {Lim}_{y \to 0^{+}} \frac{y^{2} (1 - b) - (a + b) y}{y^{'}} \frac{- 1 + y)}{1 + 0} = 0 \\ \Rightarrow {Lim}_{y \to 0^{+}} \frac{y (1 - b) - (a + b)}{1 + y} = a + b = 0 \Rightarrow b = - 1 \end{array}$