Question

$\huge \displaystyle\sf\lim\limits_{x\to\infty}\sqrt[x]{\dfrac{x!}{x^{x}}}$Or$\displaystyle\sf\lim\limits_{x\to\infty}\left(\dfrac{x!}{x}\right)^{\left(\dfrac{1}{x}\right)}$
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Answer 1

Answer:

$\large \pmb{\bf{\underline{\gray{Solution :-}}}}$

$\sf {Assume\:\:\displaystyle \lim_{x\to\infty}\left ( \dfrac{x!}{x} \right )^{\dfrac{1}{x}} = L}$

Take log both sides, we get

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( \dfrac{x!}{x} \right )}$

$\pmb{\sf{\gray{ Put\ value\ of\ x! }}}$

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( \dfrac{x(x-1)!}{x} \right )}$

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( (x-1)! \right )}$

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} \dfrac{ln \left ( (x-1)! \right )}{x}}$

$\pmb{\tt{Multiplying \ with\ ( x - 1 )\ in\ numerator\ and\ denominator\, we\ get\ }}$

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} (x-1)\dfrac{ln \left ( (x-1)! \right )}{x(x-1)}}$

$\pmb{\sf{We\ know\ that}}$

$\sf {\displaystyle \lim_{x\to\infty} \dfrac{ln \left ( x! \right )}{x}=\infty}$

$\sf {ln(L)=(\infty) \displaystyle \lim_{x\to\infty} \dfrac{x-1}{x}}$

$\sf {ln(L)=(\infty) \displaystyle \lim_{x\to\infty} \left(1-\dfrac{1}{x}\right)}$

Put value of limits,

$\sf {ln(L)=(\infty) \left(1-\dfrac{1}{\infty}\right)}$

$\sf {ln(L)=(\infty) \left(1-0\right)}$

$\sf {ln(L)=\infty}$

$\sf{L=e^{\infty}}$

$\sf{L=\infty}$

hopes its helps you..

:)