WBJEE · Maths · Limits
\(\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-a x-b\right),(a, b \in R)=0\). Then
- A \(a=0, b=1\)
- B \(a=1, b=-1\)
- C \(a=-1, b-1\)
- D \(a=0, b=0\)
Answer & Solution
Correct Answer
(B) \(a=1, b=-1\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow \infty} \frac{x^2+1-a x(x+1)-b(x+1)}{x+1}=0\) \(=\lim _{x \rightarrow \infty} \frac{(1-a) x^2-(a+b) x+1-b}{x+1}=0\) for the limit to exist,…
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