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Question Number 66444 by ~ À ® @ 237 ~ last updated on 15/Aug/19
 calculate  lim_(x→0)   (x!)^(1/x)        if     x!=Π(x)=∫_0 ^∞ t^x  e^(−t) dt
calculatelimx0(x!)1xifx!=Π(x)=0txetdt
Commented by ~ À ® @ 237 ~ last updated on 18/Aug/19
 let named it L  we know that L>0  So         ln(L)= lim_(x→0)  ln[(x!)^(1/x) ]=lim_(x→0)  ((ln(x!))/x)  as lim_(x→0)  ln(x!)= ln(Π(0))=ln(Γ(1))=ln(1)=0  ,  we can use the Hospital Rule  Then  ln(L)= lim_(x→0) (((Π^′ (x))/(Π(x)))/1) =lim_(x→0)  ((∫_0 ^∞ t^x lnt e^(−t)  dt)/(∫_0 ^∞ t^x  e^(−t)  dt )) = ((∫_0 ^∞  e^(−t) lntdt)/(∫_(0 ) ^∞ e^(−t) dt)) = ∫_0_  ^∞ e^(−t) lnt dt          because ∫_0 ^∞  e^(−t) dt = lim_(x→∞)  (1−e^(−x) )=1    Now we have   ln(L)=∫_0 ^∞  e^(−t) lnt dt = ∫_0 ^∞  lim_(n→∞)  (1−(t/n))^n lnt dt=lim_(n→∞)  J_n        with  J_n =∫_0 ^n (1−(t/n))^n lnt dt   let explicit J_(n ) in terms of n  For that , let named  u=1−(t/n)  ⇔ t=n(1−u) ⇒ dt=−ndu  J_n  = n∫_0 ^1  u^n ln(n(1−u))du= n ln(n)∫_0 ^1 u^n du +n∫_0 ^1 u^n ln(1−u)du =((nln(n))/(n+1)) + n K_n    In the way to find K_n   , we ascertain that   u^n ln(1−u)+ ((u^(n+1) −1)/((n+1)(u−1))) = (d/du)[((u^(n+1) −1)/(n+1)) ln(1−u)]  So  we have  K_n  = [((u^(n+1) −1)/(n+1)) ln(1−u)]_0 ^1 −(1/(n+1)) ∫_0 ^1 ((u^(n+1) −1)/(u−1)) du        = −(1/(n+1)) ∫_0 ^1  (1+u+u^2 +....+u^n )du       cause lim_(u→1)  (u^(n+1) −1)ln(1−u)=lim_(u→1)  −(1+u+....+u^n )(1−u)ln(1−u)=0   ( when knowing that lim_(x→0)  xlnx =0 )        =((−1)/(n+1)) Σ_(k=0) ^n  ∫_0 ^1 u^k  du = ((−1)/(n+1)) Σ_(k=0) ^n  (1/(k+1)) =((−1)/(n+1)) Σ_(k=1) ^(n+1) (1/k)   Then we reach at   J_n = (n/(n+1)) ln(n) −(n/(n+1))Σ_(k=1) ^(n+1) (1/k) = (n/(n+1))[ln(n)−H_(n+1) ]     with  H_n =Σ_(k=1) ^n (1/k)     we have  H_(n+1) = H_n  +(1/(n+1))   Using that  we get  J_n = (n/(n+1))[  ln(n) −H_n  −(1/(n+1))] = −(n/((n+1)^2 )) −(n/(n+1))[ H_n −ln(n)]  So   ln(L) =lim_(n→∞)  J_n  = lim_(n→∞)   ((−n)/((n+1)^2 )) − lim_(n→∞)  (n/(n+1)) .lim_(n→∞)  (H_n  −ln(n))= −γ      ( The Euler′s constant )  Finally        L= lim_(x→0)  (x!)^(1/x)  = e^(−γ)  .
letnameditLweknowthatL>0Soln(L)=limx0ln[(x!)1x]=limx0ln(x!)xaslimx0ln(x!)=ln(Π(0))=ln(Γ(1))=ln(1)=0,wecanusetheHospitalRuleThenln(L)=limx0(x)Π(x)1=limx00txlntetdt0txetdt=0etlntdt0etdt=0etlntdtbecause0etdt=limx(1ex)=1Nowwehaveln(L)=0etlntdt=0limn(1tn)nlntdt=limnJnwithJn=0n(1tn)nlntdtletexplicitJnintermsofnForthat,letnamedu=1tnt=n(1u)dt=nduJn=n01unln(n(1u))du=nln(n)01undu+n01unln(1u)du=nln(n)n+1+nKnInthewaytofindKn,weascertainthatunln(1u)+un+11(n+1)(u1)=ddu[un+11n+1ln(1u)]SowehaveKn=[un+11n+1ln(1u)]011n+101un+11u1du=1n+101(1+u+u2+.+un)ducauselimu1(un+11)ln(1u)=limu1(1+u+.+un)(1u)ln(1u)=0(whenknowingthatlimx0xlnx=0)=1n+1nk=001ukdu=1n+1nk=01k+1=1n+1n+1k=11kThenwereachatJn=nn+1ln(n)nn+1n+1k=11k=nn+1[ln(n)Hn+1]withHn=nk=11kwehaveHn+1=Hn+1n+1UsingthatwegetJn=nn+1[ln(n)Hn1n+1]=n(n+1)2nn+1[Hnln(n)]Soln(L)=limnJn=limnn(n+1)2limnnn+1.limn(Hnln(n))=γ(TheEulersconstant)FinallyL=limx0(x!)1x=eγ.

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