Question

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$\sf \lim_{ x \to \infty }{[ \frac{n!}{ {n}^{n} } ]}^{ {n}^{ - 1} }$
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Answer 1

Answer:-

$\sf let \ y = \displaystyle \lim_{n \to \infty} \bigg( \small \sf \frac{n!}{ {n}^{n} } \bigg)^{ \frac{1}{n} }$

$\sf \implies \log(y) = \displaystyle \small\lim_{ \sf n \to \infty} \sf \frac{1}{n} \log\bigg( \small \sf \frac{n!}{ {n}^{n} } \bigg)$

$\sf = \displaystyle \small\lim_{ \sf n \to \infty} \sf \frac{1}{n} \log\bigg( \frac{1}{n} \times \frac{2}{n} \times ... \frac{n}{n} \bigg)$

$\sf = \displaystyle \small\lim_{ \sf n \to \infty} \sf \frac{1}{n} \bigg( \log \bigg( \frac{1}{n} \bigg) + \log\bigg(\frac{2}{n} \bigg) + ... \log \bigg(\frac{n}{n} \bigg) \bigg)$

$= \displaystyle \lim_{ \sf n \to \infty} \sf \small \frac{1}{n} \sum _{r = 1}^{n} \bigg( \log \bigg( \frac{r}{n} \bigg) \bigg)$

$= \displaystyle \int_{0}^{1} \sf \small log(x) dx$

$\displaystyle = \sf \bigg( \large x \log(x) - x \bigg) ^{1}_{0}$

$\sf = 0 - 1 - \displaystyle \lim_{ \sf x \to 0^{ + } }( \sf x \log(x)) + 0$

$\sf = - 1 + 0 = - 1$

$\sf \implies \log(y) = - 1$

$\sf \implies y = {e}^{ - 1}$

$\sf \implies y =\frac{1}{e}$

This is the required answer.