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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
$$\:{calculate}\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \:\:\:\:\:\:\:{if}\:\:\:\:\:{x}!=\Pi\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{x}} \:{e}^{−{t}} {dt} \\ $$
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)  .
$$\:{let}\:{named}\:{it}\:{L} \\ $$$${we}\:{know}\:{that}\:{L}>\mathrm{0} \\ $$$${So}\:\:\:\:\:\:\:\:\:{ln}\left({L}\right)=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{ln}\left[\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \right]=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{ln}\left({x}!\right)}{{x}} \\ $$$${as}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{ln}\left({x}!\right)=\:{ln}\left(\Pi\left(\mathrm{0}\right)\right)={ln}\left(\Gamma\left(\mathrm{1}\right)\right)={ln}\left(\mathrm{1}\right)=\mathrm{0}\:\:,\:\:{we}\:{can}\:{use}\:{the}\:{Hospital}\:{Rule} \\ $$$${Then}\:\:{ln}\left({L}\right)=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\prod^{'} \left({x}\right)}{\Pi\left({x}\right)}}{\mathrm{1}}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\int_{\mathrm{0}} ^{\infty} {t}^{{x}} {lnt}\:{e}^{−{t}} \:{dt}}{\int_{\mathrm{0}} ^{\infty} {t}^{{x}} \:{e}^{−{t}} \:{dt}\:}\:=\:\frac{\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {lntdt}}{\int_{\mathrm{0}\:} ^{\infty} {e}^{−{t}} {dt}}\:=\:\int_{\mathrm{0}_{} } ^{\infty} {e}^{−{t}} {lnt}\:{dt}\:\:\:\:\:\:\:\:\:\:{because}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {dt}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}−{e}^{−{x}} \right)=\mathrm{1}\:\: \\ $$$${Now}\:{we}\:{have}\: \\ $$$${ln}\left({L}\right)=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {lnt}\:{dt}\:=\:\int_{\mathrm{0}} ^{\infty} \:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}} {lnt}\:{dt}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{J}_{{n}} \:\:\:\:\:\:\:{with}\:\:{J}_{{n}} =\int_{\mathrm{0}} ^{{n}} \left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}} {lnt}\:{dt}\: \\ $$$${let}\:{explicit}\:{J}_{{n}\:} {in}\:{terms}\:{of}\:{n} \\ $$$${For}\:{that}\:,\:{let}\:{named}\:\:{u}=\mathrm{1}−\frac{{t}}{{n}}\:\:\Leftrightarrow\:{t}={n}\left(\mathrm{1}−{u}\right)\:\Rightarrow\:{dt}=−{ndu} \\ $$$${J}_{{n}} \:=\:{n}\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}^{{n}} {ln}\left({n}\left(\mathrm{1}−{u}\right)\right){du}=\:{n}\:{ln}\left({n}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{{n}} {du}\:+{n}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{{n}} {ln}\left(\mathrm{1}−{u}\right){du}\:=\frac{{nln}\left({n}\right)}{{n}+\mathrm{1}}\:+\:{n}\:{K}_{{n}} \: \\ $$$${In}\:{the}\:{way}\:{to}\:{find}\:{K}_{{n}} \:\:,\:{we}\:{ascertain}\:{that} \\ $$$$\:{u}^{{n}} {ln}\left(\mathrm{1}−{u}\right)+\:\frac{{u}^{{n}+\mathrm{1}} −\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({u}−\mathrm{1}\right)}\:=\:\frac{{d}}{{du}}\left[\frac{{u}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}}\:{ln}\left(\mathrm{1}−{u}\right)\right] \\ $$$${So}\:\:{we}\:{have} \\ $$$${K}_{{n}} \:=\:\left[\frac{{u}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}}\:{ln}\left(\mathrm{1}−{u}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{{n}+\mathrm{1}} −\mathrm{1}}{{u}−\mathrm{1}}\:{du} \\ $$$$\:\:\:\:\:\:=\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}+{u}+{u}^{\mathrm{2}} +….+{u}^{{n}} \right){du}\:\:\:\:\:\:\:{cause}\:\underset{{u}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left({u}^{{n}+\mathrm{1}} −\mathrm{1}\right){ln}\left(\mathrm{1}−{u}\right)=\underset{{u}\rightarrow\mathrm{1}} {\mathrm{lim}}\:−\left(\mathrm{1}+{u}+….+{u}^{{n}} \right)\left(\mathrm{1}−{u}\right){ln}\left(\mathrm{1}−{u}\right)=\mathrm{0}\:\:\:\left(\:{when}\:{knowing}\:{that}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{xlnx}\:=\mathrm{0}\:\right) \\ $$$$\:\:\:\:\:\:=\frac{−\mathrm{1}}{{n}+\mathrm{1}}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{{k}} \:{du}\:=\:\frac{−\mathrm{1}}{{n}+\mathrm{1}}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\frac{−\mathrm{1}}{{n}+\mathrm{1}}\:\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}}\: \\ $$$${Then}\:{we}\:{reach}\:{at}\: \\ $$$${J}_{{n}} =\:\frac{{n}}{{n}+\mathrm{1}}\:{ln}\left({n}\right)\:−\frac{{n}}{{n}+\mathrm{1}}\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}}\:=\:\frac{{n}}{{n}+\mathrm{1}}\left[{ln}\left({n}\right)−{H}_{{n}+\mathrm{1}} \right]\:\:\:\:\:{with}\:\:{H}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:\:\:\:\:{we}\:{have}\:\:{H}_{{n}+\mathrm{1}} =\:{H}_{{n}} \:+\frac{\mathrm{1}}{{n}+\mathrm{1}}\: \\ $$$${Using}\:{that}\:\:{we}\:{get} \\ $$$${J}_{{n}} =\:\frac{{n}}{{n}+\mathrm{1}}\left[\:\:{ln}\left({n}\right)\:−{H}_{{n}} \:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right]\:=\:−\frac{{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{{n}}{{n}+\mathrm{1}}\left[\:{H}_{{n}} −{ln}\left({n}\right)\right] \\ $$$${So}\: \\ $$$${ln}\left({L}\right)\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{J}_{{n}} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{−{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:−\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}}{{n}+\mathrm{1}}\:.\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left({H}_{{n}} \:−{ln}\left({n}\right)\right)=\:−\gamma\:\:\:\:\:\:\left(\:{The}\:{Euler}'{s}\:{constant}\:\right) \\ $$$${Finally}\:\:\: \\ $$$$\:\:\:{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \:=\:{e}^{−\gamma} \:. \\ $$$$ \\ $$$$ \\ $$

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