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JEE Advanced · Mathematics · 16. Limits

If \(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4\), then

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

Answer & Solution

Correct Answer

(B) \(a=1, b=-4\)

Step-by-step Solution

Detailed explanation

Given: \(\lim _{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4\)

\(\begin{array}{l}

\Rightarrow \lim _{x \rightarrow \infty} \frac{x^{2}+x+1-a x^{2}-a x-b x-b}{x+1}=4 \\

\Rightarrow \lim _{x \rightarrow \infty} \frac{(1-a) x^{2}+(1-a-b) x+(1-b)}{x+1}=4

\end{array}\)

For this limit to be finite \(1-a=0 \Rightarrow a=1\) then given limit reduces to

\(\lim _{x \rightarrow \infty} \frac{-b x+(1-b)}{x+1}=4 \Rightarrow \lim _{x \rightarrow \infty} \frac{-b+\frac{(1-b)}{x}}{1+\frac{1}{x}}=4\)

\(\Rightarrow-b=4\) or \(b=-4, \therefore a=1, b=-4\)
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