Question Number 66444 by ~ À ® @ 237 ~ last updated on 15/Aug/19
$$\:{calculate}\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \:\:\:\:\:\:\:{if}\:\:\:\:\:{x}!=\Pi\left({x}\right)=\int_{\mathrm{0}} ^{\infty} {t}^{{x}} \:{e}^{−{t}} {dt} \\ $$
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) .](Q66660.png)
$$\:{let}\:{named}\:{it}\:{L} \\ $$$${we}\:{know}\:{that}\:{L}>\mathrm{0} \\ $$$${So}\:\:\:\:\:\:\:\:\:{ln}\left({L}\right)=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{ln}\left[\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \right]=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{ln}\left({x}!\right)}{{x}} \\ $$$${as}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{ln}\left({x}!\right)=\:{ln}\left(\Pi\left(\mathrm{0}\right)\right)={ln}\left(\Gamma\left(\mathrm{1}\right)\right)={ln}\left(\mathrm{1}\right)=\mathrm{0}\:\:,\:\:{we}\:{can}\:{use}\:{the}\:{Hospital}\:{Rule} \\ $$$${Then}\:\:{ln}\left({L}\right)=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\frac{\prod^{'} \left({x}\right)}{\Pi\left({x}\right)}}{\mathrm{1}}\:=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\int_{\mathrm{0}} ^{\infty} {t}^{{x}} {lnt}\:{e}^{−{t}} \:{dt}}{\int_{\mathrm{0}} ^{\infty} {t}^{{x}} \:{e}^{−{t}} \:{dt}\:}\:=\:\frac{\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {lntdt}}{\int_{\mathrm{0}\:} ^{\infty} {e}^{−{t}} {dt}}\:=\:\int_{\mathrm{0}_{} } ^{\infty} {e}^{−{t}} {lnt}\:{dt}\:\:\:\:\:\:\:\:\:\:{because}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {dt}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}−{e}^{−{x}} \right)=\mathrm{1}\:\: \\ $$$${Now}\:{we}\:{have}\: \\ $$$${ln}\left({L}\right)=\int_{\mathrm{0}} ^{\infty} \:{e}^{−{t}} {lnt}\:{dt}\:=\:\int_{\mathrm{0}} ^{\infty} \:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}} {lnt}\:{dt}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{J}_{{n}} \:\:\:\:\:\:\:{with}\:\:{J}_{{n}} =\int_{\mathrm{0}} ^{{n}} \left(\mathrm{1}−\frac{{t}}{{n}}\right)^{{n}} {lnt}\:{dt}\: \\ $$$${let}\:{explicit}\:{J}_{{n}\:} {in}\:{terms}\:{of}\:{n} \\ $$$${For}\:{that}\:,\:{let}\:{named}\:\:{u}=\mathrm{1}−\frac{{t}}{{n}}\:\:\Leftrightarrow\:{t}={n}\left(\mathrm{1}−{u}\right)\:\Rightarrow\:{dt}=−{ndu} \\ $$$${J}_{{n}} \:=\:{n}\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}^{{n}} {ln}\left({n}\left(\mathrm{1}−{u}\right)\right){du}=\:{n}\:{ln}\left({n}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{{n}} {du}\:+{n}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{{n}} {ln}\left(\mathrm{1}−{u}\right){du}\:=\frac{{nln}\left({n}\right)}{{n}+\mathrm{1}}\:+\:{n}\:{K}_{{n}} \: \\ $$$${In}\:{the}\:{way}\:{to}\:{find}\:{K}_{{n}} \:\:,\:{we}\:{ascertain}\:{that} \\ $$$$\:{u}^{{n}} {ln}\left(\mathrm{1}−{u}\right)+\:\frac{{u}^{{n}+\mathrm{1}} −\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({u}−\mathrm{1}\right)}\:=\:\frac{{d}}{{du}}\left[\frac{{u}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}}\:{ln}\left(\mathrm{1}−{u}\right)\right] \\ $$$${So}\:\:{we}\:{have} \\ $$$${K}_{{n}} \:=\:\left[\frac{{u}^{{n}+\mathrm{1}} −\mathrm{1}}{{n}+\mathrm{1}}\:{ln}\left(\mathrm{1}−{u}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{u}^{{n}+\mathrm{1}} −\mathrm{1}}{{u}−\mathrm{1}}\:{du} \\ $$$$\:\:\:\:\:\:=\:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}+{u}+{u}^{\mathrm{2}} +....+{u}^{{n}} \right){du}\:\:\:\:\:\:\:{cause}\:\underset{{u}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\left({u}^{{n}+\mathrm{1}} −\mathrm{1}\right){ln}\left(\mathrm{1}−{u}\right)=\underset{{u}\rightarrow\mathrm{1}} {\mathrm{lim}}\:−\left(\mathrm{1}+{u}+....+{u}^{{n}} \right)\left(\mathrm{1}−{u}\right){ln}\left(\mathrm{1}−{u}\right)=\mathrm{0}\:\:\:\left(\:{when}\:{knowing}\:{that}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{xlnx}\:=\mathrm{0}\:\right) \\ $$$$\:\:\:\:\:\:=\frac{−\mathrm{1}}{{n}+\mathrm{1}}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{{k}} \:{du}\:=\:\frac{−\mathrm{1}}{{n}+\mathrm{1}}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{{k}+\mathrm{1}}\:=\frac{−\mathrm{1}}{{n}+\mathrm{1}}\:\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}}\: \\ $$$${Then}\:{we}\:{reach}\:{at}\: \\ $$$${J}_{{n}} =\:\frac{{n}}{{n}+\mathrm{1}}\:{ln}\left({n}\right)\:−\frac{{n}}{{n}+\mathrm{1}}\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}}\:=\:\frac{{n}}{{n}+\mathrm{1}}\left[{ln}\left({n}\right)−{H}_{{n}+\mathrm{1}} \right]\:\:\:\:\:{with}\:\:{H}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\:\:\:\:\:{we}\:{have}\:\:{H}_{{n}+\mathrm{1}} =\:{H}_{{n}} \:+\frac{\mathrm{1}}{{n}+\mathrm{1}}\: \\ $$$${Using}\:{that}\:\:{we}\:{get} \\ $$$${J}_{{n}} =\:\frac{{n}}{{n}+\mathrm{1}}\left[\:\:{ln}\left({n}\right)\:−{H}_{{n}} \:−\frac{\mathrm{1}}{{n}+\mathrm{1}}\right]\:=\:−\frac{{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:−\frac{{n}}{{n}+\mathrm{1}}\left[\:{H}_{{n}} −{ln}\left({n}\right)\right] \\ $$$${So}\: \\ $$$${ln}\left({L}\right)\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{J}_{{n}} \:=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{−{n}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\:−\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}}{{n}+\mathrm{1}}\:.\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left({H}_{{n}} \:−{ln}\left({n}\right)\right)=\:−\gamma\:\:\:\:\:\:\left(\:{The}\:{Euler}'{s}\:{constant}\:\right) \\ $$$${Finally}\:\:\: \\ $$$$\:\:\:{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left({x}!\right)^{\frac{\mathrm{1}}{{x}}} \:=\:{e}^{−\gamma} \:. \\ $$$$ \\ $$$$ \\ $$