Question

Lagyan ng tsek ang patlang kung ang pangungusap ay nagsasood
ng pogpopodlogo otekis [ X ] kung hindi.
1. Maraming kababaihan ang tumulong sa maraming paraan sa rebolusyon at ang ilan sa kanic ay sina
Melchora Aquino, Gabriela Silang, Teresa Magbanua, Gliceria, at Patrocinio
2. Maraming Pilipino ang nagbuwis ng kanilang buhay para sa kanilang mga ipinaglalabon
3. Kinamkam ng mga prayle ang mga lupain ng mga katutubo.
4. Pagtatatag ng ihim na samahan para bumuo ng pag aaklas laban sa mga dayuhan
5. May mga katutubong sumorib sa mga Espanyol at nücbanan ang kapwa
Pilipino sa panahon ng pag-aalsa










*wag mag coment pag hindi alam*"​

Answer 1

$\large \pmb{\bf{\underline{\gray{Solution :-}}}}$

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

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,

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

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

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

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

$\sf{L=\infty}$

Answer 2

question :

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

$\large \pmb{\bf{\underline{\gray{Solution :-}}}}$

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

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,

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

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

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

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

$\sf{L=\infty}$