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lim-x-x-x-x-1-x-




Question Number 80748 by jagoll last updated on 06/Feb/20
lim_(x→∞)  (((x!)/x^x ))^(1/x)  = ?
limx(x!xx)1x=?
Commented by john santu last updated on 06/Feb/20
let y = lim_(x→∞ ) (((x!)/x^x ))^(1/x)   ln(y)=lim_(x→∞ ) ln(((x!)/x^x ))^(1/x)   ln(y)=lim_(x→∞)  ln(((x!)/x^x ))^(1/x)   ln(y)=lim_(x→∞) (1/x)ln(((x!)/x^x ))  = lim_(x→∞) (1/x)(ln((1/x))+ln((2/x))+...+ln(((x−1)/x))+ln((x/x)))  = ∫_0 ^1 ln(x)dx = −1  ln(y)= −1 ⇒ ∴ y = e^(−1)   lim_(x→∞)  (((x!)/x^x ))^(1/x) = (1/e)
lety=limx(x!xx)1xln(y)=limxln(x!xx)1xln(y)=limxln(x!xx)1xln(y)=limx1xln(x!xx)=limx1x(ln(1x)+ln(2x)++ln(x1x)+ln(xx))=01ln(x)dx=1ln(y)=1y=e1limx(x!xx)1x=1e
Commented by jagoll last updated on 06/Feb/20
thanks
thanks
Commented by mathmax by abdo last updated on 06/Feb/20
let f(x)=(((x!)/x^x ))^(1/x)   we have x! ∼x^x  e^(−x) (√(2πx))(x→+∞) ⇒  f(x)∼(e^(−x) (√(2πx)))^(1/x) =e^((1/x)ln(e^(−x) (√(2πx))))  =e^(−1 +(1/(2x))ln(2πx))  ⇒e^(−1) =(1/e)(x→+∞)  ⇒lim_(x→+∞)  f(x)=(1/e)
letf(x)=(x!xx)1xwehavex!xxex2πx(x+)f(x)(ex2πx)1x=e1xln(ex2πx)=e1+12xln(2πx)e1=1e(x+)limx+f(x)=1e
Answered by MJS last updated on 06/Feb/20
lim_(x→∞)  (((x!)/x^x ))^(1/x)  =       [x!≈(x^x /e^x )(√(2πx))  Sterling′s approximation]  =(1/e)lim_(x→∞)  (2πx)^(1/(2x))  =(1/e)       [lim_(x→∞)  C_1 ^(1/x) =1 ∀C_1 >0]       [lim_(x→∞)  x^(1/(C_2 x))  =lim_(x→∞)  e^((ln  x)/(C_2 x)) =10^0 =1 ∀C_2 ≠0]
limx(x!xx)1x=[x!xxex2πxSterlingsapproximation]=1elimx(2πx)12x=1e[limxC11x=1C1>0][limxx1C2x=limxelnxC2x=100=1C20]
Commented by jagoll last updated on 06/Feb/20
mister the answer (1/e) mister  what wrong?
mistertheanswer1emisterwhatwrong?
Commented by MJS last updated on 06/Feb/20
the answer is there in line 3
theansweristhereinline3
Commented by jagoll last updated on 06/Feb/20
oo thank you mister. i see the  last line
oothankyoumister.iseethelastline

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