Math Question and Answers


Question and Answers Forum

All Questions Topic List
Limits Questions
Previous in All Question Next in All Question
Previous in Limits Next in Limits
Question Number 60981 by necx1 last updated on 28/May/19

Commented by Prithwish sen last updated on 28/May/19

$Let$ $A = \lim_{x \to \infty} {(\frac{x!}{x^{x}})}^{\frac{1}{x}}$ $∴ lnA = \lim_{x \to \infty} \frac{1}{x} \ln (\frac{1.2.3 . . . . . . . . . . . x}{x . x . x . . . . . . . . . . . . x})$ $= \lim_{x \to \infty} \frac{1}{x} (\ln \frac{1}{x} + \ln \frac{2}{x} + . . . . . . . . \ln \frac{x}{x})$ $= \lim_{x \to \infty} \frac{1}{x} \underset{r = 1}{\sum^{x}} \ln \frac{r}{x}$ $Missing \left or extra \right$ $= - 1$ $∴ A = e^{- 1} = \frac{1}{e}$
Commented by tanmay last updated on 28/May/19

$b a h d a r u n e x c e l l e n t . . .$
Commented by Prithwish sen last updated on 28/May/19

$thank you sir .$
Commented by ajfour last updated on 28/May/19

$E x t r e m e l y g r e a t S i r, t h a n k s a l o t .$
Commented by Prithwish sen last updated on 28/May/19

$welcome .$
Commented by Mr X pcx last updated on 28/May/19
$stirling gormulae n! ∼ n^n e^(−n) (√(2πn)) or x! ∼x^x e^(−x) (√(2πx)) ⇒((x!)/x^x ) ∼ e^(−x) (√(2πx)) ⇒ (((x!)/x^x ))^(1/x) =(e^(−x) (√(2πx)))^(1/x) =A(x) ln(A(x)) =(1/x)ln(e^(−x) (√(2πx))) =−1 +(1/(2x)){ln(2π) +ln(x)} =−1 +((ln(2π))/(2x)) +((ln(x))/(2x)) →−1(x→+∞) ⇒lim_(x→+∞) A(x)=e^(−1) =(1/e) so lim_(x→+∞) (((x!)/x^x ))^(1/x) =(1/e) .$
$s t i r l i n g g o r m u l a e$ $n! \sim n^{n} e^{- n} \sqrt{2 π n} o r$ $x! \sim x^{x} e^{- x} \sqrt{2 π x} \Rightarrow \frac{x!}{x^{x}} \sim e^{- x} \sqrt{2 π x} \Rightarrow$ ${(\frac{x!}{x^{x}})}^{\frac{1}{x}} = {(e^{- x} \sqrt{2 π x})}^{\frac{1}{x}} = A (x)$ $l n (A (x)) = \frac{1}{x} l n (e^{- x} \sqrt{2 π x})$ $= - 1 + \frac{1}{2 x} {l n (2 π) + l n (x)}$ $= - 1 + \frac{l n (2 π)}{2 x} + \frac{l n (x)}{2 x} \to - 1 (x \to + \infty)$ $\Rightarrow {l i m}_{x \to + \infty} A (x) = e^{- 1} = \frac{1}{e}$ $s o {l i m}_{x \to + \infty} {(\frac{x!}{x^{x}})}^{\frac{1}{x}} = \frac{1}{e} .$
	Answered by MJS last updated on 28/May/19

	$tried with {Stirling}^{'} s formula$ $x! \approx {(\frac{x}{e})}^{x} \sqrt{2 π x}$ ${(\frac{x!}{x^{x}})}^{\frac{1}{x}} = \frac{x!^{\frac{1}{x}}}{x} \approx \frac{1}{x} {({(\frac{x}{e})}^{x} \sqrt{2 π x})}^{\frac{1}{x}} = \frac{1}{x} \times \frac{x}{e} {(2 π x)}^{\frac{1}{2 x}} =$ $= \frac{1}{e} \times 2^{\frac{1}{2 x}} \times π^{\frac{1}{2 x}} \times x^{\frac{1}{2 x}}$ $\lim_{x \to \infty} \frac{1}{e} \times 2^{\frac{1}{2 x}} \times π^{\frac{1}{2 x}} \times x^{\frac{1}{2 x}} = \frac{1}{e}$ $this seems right, computing shows$ ${(\frac{100!}{100^{100}})}^{\frac{1}{100}} \approx . 379927$ ${(\frac{200!}{200^{200}})}^{\frac{1}{200}} \approx . 374502$ ${(\frac{300!}{300^{300}})}^{\frac{1}{300}} \approx . 372533$ ${(\frac{400!}{400^{400}})}^{\frac{1}{400}} \approx . 371498$ $\frac{1}{e} \approx . 367879$
	Commented by tanmay last updated on 28/May/19

	$b a h d a r u n e x c e l l e n t . . . o n e p r o b l e m b u t s o l u t i o n$ $t h r e e m e t h o d . . .$
	Answered by tanmay last updated on 28/May/19
	$lim_(x→∞) (a_(x+1) /a_x ) =lim_(x→∞) (({(((x+1)!)/((x+1)^(x+1) ))})/((((x!)/x^x )))) =lim_(x→∞) (((x+1)!)/(x!))×(1/(((x+1)^(x+1) )/x^x )) =lim_(x→∞) (x+1)×(1/((x+1)(((x+1)/x))^x )) =lim_(x→∞) (1/((1+(1/x))^x )) now t=(1/x) y=lim_(t→0) (1/((1+t)^(1/t) )) lny=lim_(t→0) [((−ln(1+t))/t)] lny=−1 y=e^(−1) so lim_(x→∞) (((x!)/x^x ))^(1/x) =e^(−1) →it is the answer using cauchy second theorem on limit lim_(n→∞) (a_n )^(1/n) =l if lim_(n→∞) (a_(n+1) /a_n )=l$
	$\lim_{x \to \infty} \frac{a_{x + 1}}{a_{x}}$ $= \lim_{x \to \infty} \frac{{\frac{(x + 1)!}{{(x + 1)}^{x + 1}}}}{(\frac{x!}{x^{x}})}$ $= \lim_{x \to \infty} \frac{(x + 1)!}{x!} \times \frac{1}{\frac{{(x + 1)}^{x + 1}}{x^{x}}}$ $= \lim_{x \to \infty} (x + 1) \times \frac{1}{(x + 1) {(\frac{x + 1}{x})}^{x}}$ $= \lim_{x \to \infty} \frac{1}{{(1 + \frac{1}{x})}^{x}}$ $n o w t = \frac{1}{x}$ $y = \lim_{t \to 0} \frac{1}{{(1 + t)}^{\frac{1}{t}}}$ $l n y = \lim_{t \to 0} [\frac{- l n (1 + t)}{t}]$ $l n y = - 1$ $y = e^{- 1}$ $s o \lim_{x \to \infty} {(\frac{x!}{x^{x}})}^{\frac{1}{x}} = e^{- 1} \to i t i s t h e a n s w e r$ $u s i n g c a u c h y s e c o n d t h e o r e m o n l i m i t$ $\lim_{n \to \infty} {(a_{n})}^{\frac{1}{n}} = l$ $i f \lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = l$
	Commented by necx1 last updated on 28/May/19

	$G o d b l e s s y o u s i r$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum