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AP EAMCET · Maths · Limits

If \(\lim _{x \rightarrow \infty}\left\{\frac{x^3+1}{x^2+1}-(\alpha x+\beta)\right\}\) exists and equal to 2 , then the ordered pair \((\alpha, \beta)\) of real numbers is

  1. A \((1,-1)\)
  2. B \((-2,1)\)
  3. C \((-1,1)\)
  4. D \((1,-2)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \((1,-2)\)

Step-by-step Solution

Detailed explanation

It is given that, \(\begin{aligned} & \lim _{x \rightarrow \infty}\left\{\frac{x^3+1}{x^2+1}-(\alpha x+\beta)\right\}=2 \\ & \Rightarrow \lim _{x \rightarrow \infty} \frac{x^3+1-\alpha x^3-\beta x^2-\alpha x-\beta}{x^2+1}=2 \end{aligned}\) For the existance of limit, coefficient…