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$\lim _{n \rightarrow \infty} \frac{1}{n}\left(\frac{1}{e^{1 / n}}+\frac{1}{e^{2 / n}}+\frac{1}{e^{3 / n}}+\ldots+\frac{1}{e^2}\right)=$
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The correct answer is:
$1-\mathrm{e}^{-2}$
$\lim _{h \rightarrow \infty} \frac{1}{n}\left(\frac{1}{e^{1 / n}}+\frac{1}{e^{2 / n}}+\frac{1}{e^{3 / n}}+\ldots .+\frac{1}{e^2}\right)$
$=\lim _{h \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{2 n} \frac{1}{e^{k / n}}$
$\begin{aligned} & =\int_0^2 \frac{1}{e^x} d x=\int_0^2 e^{-x} d x=\left[-e^{-x}\right]_0^2 \\ & =-e^{-2}+e^0=1-e^{-2}\end{aligned}$
$=\lim _{h \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{2 n} \frac{1}{e^{k / n}}$
$\begin{aligned} & =\int_0^2 \frac{1}{e^x} d x=\int_0^2 e^{-x} d x=\left[-e^{-x}\right]_0^2 \\ & =-e^{-2}+e^0=1-e^{-2}\end{aligned}$
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