The value of lim _{n rightarrow infty} frac{( | vedclass

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(D) Let $L = \lim _{n \rightarrow \infty} \frac{(n !)^{1 / n}}{n} = \lim _{n \rightarrow \infty} \left( \frac{n!}{n^n} \right)^{1/n}$.Taking the natur

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$\frac{1}{e}$

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(D) Let $L = \lim _{n \rightarrow \infty} \frac{(n !)^{1 / n}}{n} = \lim _{n \rightarrow \infty} \left( \frac{n!}{n^n} \right)^{1/n}$.Taking the natural logarithm on both sides:$\ln L = \lim _{n \rightarrow \infty} \frac{1}{n} \ln \left( \frac{n!}{n^n} \right) = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \ln \left( \frac{r}{n} \right)$.This is a Riemann sum for the integral $\int_{0}^{1} \ln(x) \, dx$.$\ln L = \int_{0}^{1} \ln(x) \, dx = [x \ln x - x]_{0}^{1}$.Evaluating the limit as $x \rightarrow 0^+$,we get $\lim_{x \rightarrow 0^+} x \ln x = 0$.So,$\ln L = (1 \ln 1 - 1) - (0) = -1$.Therefore,$L = e^{-1} = \frac{1}{e}$.

The value of $\lim _{n \rightarrow \infty} \frac{(n !)^{1 / n}}{n}$ is

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