Question

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

Answer 1

Topic :-

Limits

To Solve :-

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

Solution :-

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

Take log both sides,

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

x! = x( x - 1 )( x - 2 ). . . . . . . 3 × 2 × 1

x! = x( x - 1 )!

Put value of x!,

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

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} \dfrac{ln \left ( (x-1)! \right )}{x}}ln(L)= </p><p>x→∞</p><p>lim$

Multiplying with ( x - 1 ) in numerator and denominator,

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

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,

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

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

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

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

$\sf{L=\infty}$

Answer :-

So, value of the limit 'L' is

$\bold {\infty}$

Answer 2

$\huge \sf \underline \red{answer}$

To Solve :-

$\sf {\displaystyle \lim_{x\to\infty}\left ( \dfrac{x!}{x} \right )^{\dfrac{1}{x}}}x→∞lim(xx!)x1$

Solution :-

$\sf {Assume\:\:\displaystyle \lim_{x\to\infty}\left ( \dfrac{x!}{x} \right )^{\dfrac{1}{x}} = L}Assumex→∞lim(xx!)x1=L$

Take log both sides,

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( \dfrac{x!}{x} \right )}ln(L)=x→∞lim(x1)ln(xx!)</p><p>x! = x( x - 1 )( x - 2 ). . . . . . . 3 × 2 × 1</p><p>x! = x( x - 1 )!$

Put value of x!,

$</p><p>\sf {ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( \dfrac{x(x-1)!}{x} \right )} < /p > < p > \sf {ln(L)=\displaystyle \lim_{x\to\infty} \left ( \dfrac{1}{x} \right )ln \left ( (x-1)! \right )}ln(L)=x→∞lim(x1)ln(xx(x−1)!)</p><p>ln(L)=x→∞lim(x1)ln((x−1)!)$

$\sf {ln(L)=\displaystyle \lim_{x\to\infty} \dfrac{ln \left ( (x-1)! \right )}{x}}ln(L)= < /p > < p > x→∞ < /p > < p > limln(L)=x→∞limxln((x−1)!)ln(L)=</p><p>x→∞</p><p>lim$

$Multiplying \: with ( x - 1 ) in \: numerator \: and \: denominator,</p><p>\sf {ln(L)=\displaystyle \lim_{x\to\infty} (x-1)\dfrac{ln \left ( (x-1)! \right )}{x(x-1)}}ln(L)=x→∞lim(x−1)x(x−1)ln((x−1)!)$

We know that,

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

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

$\sf {ln(L)=(\infty) \displaystyle \lim_{x\to\infty} \left(1-\dfrac{1}{x}\right)}ln(L)=(∞)x→∞lim(1−x1)$

Put value of limits,

$\sf {ln(L)=(\infty) \left(1-\dfrac{1}{\infty}\right)}ln(L)=(∞)(1−∞1)</p><p>\sf {ln(L)=(\infty) \left(1-0\right)}ln(L)=(∞)(1−0)$

$\sf {ln(L)=\infty}ln(L)=∞</p><p>\sf{L=e^{\infty}}L=e∞</p><p>\sf{L=\infty}L=∞$

Answer :-

$So, \: value \: of \: the \: limit \: 'L' is \: </p><p>\bold {\infty}∞$