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

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

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,

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

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

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

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

$\sf{L=\infty}$

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Answer 2

Answer:

Assume

x→∞

lim

(

x

x!

)

x

1

=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 )}ln(L)=

x→∞

lim

(

x

1

)ln(

x

x!

)

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

x→∞

lim

(

x

1

)ln(

x

x(x−1)!

)

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

x→∞

lim

(

x

1

)ln((x−1)!)

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

x→∞

lim

x

ln((x−1)!)

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

x→∞

lim

(x−1)

x(x−1)

ln((x−1)!)

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

x→∞

lim

x

ln(x!)

=∞

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

x→∞

lim

x

x−1

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

x→∞

lim

(1−

x

1

)

Put value of limits,

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

∞

1

)

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

\sf {ln(L)=\infty}ln(L)=∞

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

∞

\sf{L=\infty}L=∞