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Question: The values of constants ‘a’ and ‘b’ so that \(\lim_{x \rightarrow \infty}\left( \frac{x^{2} - 1}{x ...

The values of constants ‘a’ and ‘b’ so that

lim⁡x→∞(x2−1x+1−ax−b)=2\lim_{x \rightarrow \infty}\left( \frac{x^{2} - 1}{x + 1} - ax - b \right) = 2 is

A

a=0,b=0a = 0,b = 0

B

a=1,b=−1a = 1,b = - 1

C

a=1,b=−3a = 1,b = - 3

D

a=2,b=−1a = 2,b = - 1

Answer

a=1,b=−3a = 1,b = - 3

Explanation

Solution

lim⁡x→∞(x2−1x+1−ax−b)=2\lim_{x \rightarrow \infty}\left( \frac{x^{2} - 1}{x + 1} - ax - b \right) = 2

⇒lim⁡x→∞x−1−ax−b=2⇒lim⁡x→∞x(1−a)−(1+b)=2\Rightarrow \lim_{x \rightarrow \infty}x - 1 - ax - b = 2 \Rightarrow \lim_{x \rightarrow \infty}x(1 - a) - (1 + b) = 2.

Comparing the coefficient of both sides, 1−a=01 - a = 0

and 1+b=−2⇒a=1,b=−31 + b = - 2 \Rightarrow a = 1,b = - 3