AP EAMCET · Maths · Limits
\(\lim _{n \rightarrow \infty} \frac{1}{n}\left(\frac{1}{e^{1 / n}}+\frac{1}{e^{2 / n}}+\frac{1}{e^{3 / n}}+\ldots+\frac{1}{e^2}\right)=\)
- A \(1-\mathrm{e}^{-2}\)
- B \(1+\mathrm{e}^{-2}\)
- C \(\mathrm{e}^2-1\)
- D \(\mathrm{e}^2+1\)
Answer & Solution
Correct Answer
(A) \(1-\mathrm{e}^{-2}\)
Step-by-step Solution
Detailed explanation
\(\lim _{h \rightarrow \infty} \frac{1}{n}\left(\frac{1}{e^{1 / n}}+\frac{1}{e^{2 / n}}+\frac{1}{e^{3 / n}}+\ldots .+\frac{1}{e^2}\right)\) \(=\lim _{h \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{2 n} \frac{1}{e^{k / n}}\)…
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