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Answer fast guys plz I knew so easy question but I forgot some steps how to solvex⁴+2x³-2x²+2x-3 is divisible byx²+2x-3.​

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Answer:

\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)}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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