Math, asked by ptsbankxd, 2 days ago

\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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Answered by Anonymous
16

Answer:

 <marquee behaviour-move><font color="blue black"><h2>here is ur answer </ ht></marquee>

\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..

Answered by RainCloud
0

Answer:

refer to the attachment

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