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Question Number 54566 by mr W last updated on 06/Feb/19

lim_(n→∞)  (((1+c)/(1+μ^(1/n) )))^n =?  with 0≤c≤1, μ>0

limn(1+c1+μ1n)n=? with0c1,μ>0

Commented bymaxmathsup by imad last updated on 06/Feb/19

let A_n =(((1+e)/(1+μ^(1/n) )))^n  ⇒  A_n =(((1+e +μ^(1/n) −μ^(1/n) )/(1+μ^(1/n) )))^n  =(1+((e−μ^(1/n) )/(1+μ^(1/n) )))^n   =e^(nln(1+((e−μ^(1/n) )/(1+μ^(1/n) )))    but  μ^(1/n)  =e^((ln(μ))/n)    ∼1+((ln(μ))/n))  ⇒e−μ^(1/n)  ∼e−1−((ln(μ))/n)  and  1+μ^(1/n)  =2 +((ln(μ))/n) ⇒ 1+((e−μ^(1/n) )/(1+μ^(1/n) )) ∼ 1 +((e−1)/2) =((1+e)/2) ⇒A_n ∼e^(nln(((1+e)/2)))  →+∞(n→+∞)  (look that ln(((1+e)/2))>0 )  we consider μ and e fixed ..

letAn=(1+e1+μ1n)nAn=(1+e+μ1nμ1n1+μ1n)n=(1+eμ1n1+μ1n)n =enln(1+eμ1n1+μ1n)butμ1n=eln(μ)n1+ln(μ)neμ1ne1ln(μ)nand 1+μ1n=2+ln(μ)n1+eμ1n1+μ1n1+e12=1+e2Anenln(1+e2)+(n+) (lookthatln(1+e2)>0)weconsiderμandefixed..

Commented bymr W last updated on 06/Feb/19

thank you sir!  c≤1  therefore ln (((1+c)/2))≤0  it seems if c<1, (((1+e)/(1+μ^(1/n) )))^n →0  what if c=1?, i.e. ((2/(1+μ^(1/n) )))^n →?

thankyousir! c1 thereforeln(1+c2)0 itseemsifc<1,(1+e1+μ1n)n0 whatifc=1?,i.e.(21+μ1n)n?

Commented bymaxmathsup by imad last updated on 06/Feb/19

you are right sir  i have commited a error  of calculus we have   0<e<1 ⇒1<1+e<2 ⇒(1/2)<((1+e)/2)<1 ⇒ln(((1+e)/2))<0 ⇒lim_(n→+∞)    A_n =0   forgive me i was tired...

youarerightsirihavecommitedaerrorofcalculuswehave 0<e<11<1+e<212<1+e2<1ln(1+e2)<0limn+An=0 forgivemeiwastired...

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