If limx→∞(x2−1x+1−ax−b) =2, then

limx→∞(x2−1−ax2−ax−bx−bx+1)=2⇒limx→∞((1−a)x2−(a+b)x−(1+b)x+1)=2 a = 1, b + 1 = -2 ⇒b = -3

If limx→∞(x2−1x+1−ax−b) =2, then

limx→∞(x2−1−ax2−ax−bx−bx+1)=2⇒limx→∞((1−a)x2−(a+b)x−(1+b)x+1)=2 a = 1, b + 1 = -2 ⇒b = -3

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If $\underset{x \to \infty}{l i m} (\frac{x^{2} - 1}{x + 1} - a x - b)$ =2, then

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a =1 and b = - 3

a = 1 and b = 2

a = 0 and b = -1

a = 2 and b = 1

answer is A.

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$\underset{x \to \infty}{l i m} (\frac{x^{2} - 1 - a x^{2} - a x - b x - b}{x + 1}) = 2$ $\Rightarrow \lim_{x \to \infty} (\frac{(1 - a) x^{2} - (a + b) x - (1 + b)}{x + 1}) = 2$
a = 1, b + 1 = -2 $\Rightarrow$ b = -3

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If limx→∞(x2−1x+1−ax−b) =2, then

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If $\underset{x \to \infty}{l i m} (\frac{x^{2} - 1}{x + 1} - a x - b)$ =2, then