MHT CET · Maths · Limits
\(\lim _{n \rightarrow \infty} \frac{1}{n^3+1}+\frac{4}{n^3+1}+\frac{9}{n^3+1}+\ldots \ldots \ldots \ldots+\frac{n^2}{n^3+1}=\)
- A \(\frac{1}{2}\)
- B \(\frac{1}{3}\)
- C \(\frac{1}{6}\)
- D \(\frac{1}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
\(\lim _{n \rightarrow \infty} \frac{1^2+2^2+\ldots+n^2}{n^3+1}\) \(= \lim _{n \rightarrow \infty} \frac{\frac{n(n+1)(2n+1)}{6}}{n^3+1}\)
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