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

Answer:

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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!,

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

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

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

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 $\bold {\infty}$.

Explanation:

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Answer 2

Answer:

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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!,

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

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

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

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 $\bold {\infty}$.

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