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Question-106662




Question Number 106662 by mathdave last updated on 06/Aug/20
Answered by Ar Brandon last updated on 06/Aug/20
Let A_n =lim_(x→0)  ln(x!)^(1/x)  , x∈N  lnx!=ln((x)(x−1)(x−2)...(x−(x−1)))           =lnΠ_(k=0) ^(x−1) (x−k)=Σ_(k=0) ^(x−1) ln(x−k)  A_n =lim_(x→0) {(1/x)Σ_(k=0) ^(x−1) ln(x−k)} ?????    However, if x→∞ we′ll have;  A_n =lim_(x→∞) {(1/x)Σ_(k=0) ^(x−1) ln(x−k)}=lim_(x→∞) ∫_0 ^1 lnx(1−t)dt        =lim_(x→∞) ∫_0 ^1 ln(ux)du=lim_(x→∞) {((ux)/x)(ln(ux)−1)}_0 ^1         =lim_(x→∞) {(lnx−1)−lim_(u→0)  [uln(ux)−u]}        =lim_(x→∞) {lnx−1}=+∞
$$\mathrm{Let}\:\mathrm{A}_{\mathrm{n}} =\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\mathrm{ln}\left(\mathrm{x}!\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:,\:\mathrm{x}\in\mathbb{N} \\ $$$$\mathrm{lnx}!=\mathrm{ln}\left(\left(\mathrm{x}\right)\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)…\left(\mathrm{x}−\left(\mathrm{x}−\mathrm{1}\right)\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{ln}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{x}−\mathrm{1}} {\prod}}\left(\mathrm{x}−\mathrm{k}\right)=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{x}−\mathrm{1}} {\sum}}\mathrm{ln}\left(\mathrm{x}−\mathrm{k}\right) \\ $$$$\mathrm{A}_{\mathrm{n}} =\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{x}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{x}−\mathrm{1}} {\sum}}\mathrm{ln}\left(\mathrm{x}−\mathrm{k}\right)\right\}\:????? \\ $$$$ \\ $$$$\mathrm{However},\:\mathrm{if}\:\mathrm{x}\rightarrow\infty\:\mathrm{we}'\mathrm{ll}\:\mathrm{have}; \\ $$$$\mathrm{A}_{\mathrm{n}} =\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{x}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{x}−\mathrm{1}} {\sum}}\mathrm{ln}\left(\mathrm{x}−\mathrm{k}\right)\right\}=\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{lnx}\left(\mathrm{1}−\mathrm{t}\right)\mathrm{dt} \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{ux}\right)\mathrm{du}=\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{ux}}{\mathrm{x}}\left(\mathrm{ln}\left(\mathrm{ux}\right)−\mathrm{1}\right)\right\}_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left\{\left(\mathrm{lnx}−\mathrm{1}\right)−\underset{\mathrm{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\mathrm{uln}\left(\mathrm{ux}\right)−\mathrm{u}\right]\right\} \\ $$$$\:\:\:\:\:\:=\underset{\mathrm{x}\rightarrow\infty} {\mathrm{lim}}\left\{\mathrm{lnx}−\mathrm{1}\right\}=+\infty \\ $$
Commented by mathdave last updated on 07/Aug/20
no d answer is=1
$${no}\:{d}\:{answer}\:{is}=\mathrm{1} \\ $$$$ \\ $$

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