Solveeit Logo
Login

Question

Question: If \(\lim_{x \rightarrow \infty}\left\lbrack \frac{x^{3} + 1}{x^{2} + 1} - (ax + b) \right\rbrack = ...

If lim⁡x→∞[x3+1x2+1−(ax+b)]=2,\lim_{x \rightarrow \infty}\left\lbrack \frac{x^{3} + 1}{x^{2} + 1} - (ax + b) \right\rbrack = 2, then

A

a=1a = 1andb=1b = 1

B

a=1a = 1 and b=−1b = - 1

C

a=1a = 1andb=−2b = - 2

D

a=1a = 1 and b=2b = 2

Answer

a=1a = 1andb=−2b = - 2

Explanation

Solution

limx→∞(x3+1x2+1−(ax+b))=2\underset{x \rightarrow \infty}{\text{lim}}{}\left( \frac{x^{3} + 1}{x^{2} + 1} - (ax + b) \right) = 2

⇒limx→∞(x3(1−a)−bx2−ax+(1−b)x2+1)=2\underset{x \rightarrow \infty}{\text{lim}}{}\left( \frac{x^{3}(1 - a) - bx^{2} - ax + (1 - b)}{x^{2} + 1} \right) = 2

⇒ lim⁡x→∞[x3(1−a)−bx2−ax+(1−b)]=2(x2+1)\lim_{x \rightarrow \infty}\lbrack x^{3}(1 - a) - bx^{2} - ax + (1 - b)\rbrack = 2(x^{2} + 1).

Comparing the coefficients of both sides, 1−a=01 - a = 0 and −b=2- b = 2 or a=1,b=−2a = 1,b = - 2.