Question

Write a letter to your younger brother advising him not to adopt unfair means in
the examination. Also guide him how to study in a better way so that he can score
more marks. (Word limit 80 – 100 words)

Answer 1

Explanation:

\large \pmb{\bf{\underline{\gray{Solution :-}}}}Solution:−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, 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(x1)ln(xx!)

\pmb{\sf{\gray{ Put\ value\ of\ x! }}}Put value of x!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 )}ln(L)=x→∞lim(x1)ln(xx(x−1)!)

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

\pmb{\tt{Multiplying \ with\ ( x - 1 )\ in\ numerator\ and\ denominator\, we\ get\ }}Multiplying with (x−1) in numerator and denominatorwe get Multiplying with (x−1) in numerator and denominatorwe get 

\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)!)

\pmb{\sf{We\ know\ that}}We know thatWe 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)

\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=∞