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Question Number 157216 by Khalmohmmad last updated on 21/Oct/21

lim_(x→∞)  (((n!)/n^n ))^(1/n) =?  lim_(x→0)  ((√(tan 4x))/(tan(√(4x))))=?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} =? \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{tan}\:\mathrm{4}{x}}}{\mathrm{tan}\sqrt{\mathrm{4}{x}}}=? \\ $$

Commented by cortano last updated on 21/Oct/21

 (2)lim_(x→0^+ )  ((√(tan 4x))/(tan (√(4x)))) = lim_(x→0^+ )  (((√(tan 4x))/( (√(4x))))/((tan (√(4x)))/( (√(4x)))))   = lim_(x→0^+ )  ((√((tan 4x)/(4x)))/((tan (√(4x)))/( (√(4x))))) =1

$$\:\left(\mathrm{2}\right)\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\sqrt{\mathrm{tan}\:\mathrm{4}{x}}}{\mathrm{tan}\:\sqrt{\mathrm{4}{x}}}\:=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\frac{\sqrt{\mathrm{tan}\:\mathrm{4}{x}}}{\:\sqrt{\mathrm{4}{x}}}}{\frac{\mathrm{tan}\:\sqrt{\mathrm{4}{x}}}{\:\sqrt{\mathrm{4}{x}}}} \\ $$$$\:=\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{\sqrt{\frac{\mathrm{tan}\:\mathrm{4}{x}}{\mathrm{4}{x}}}}{\frac{\mathrm{tan}\:\sqrt{\mathrm{4}{x}}}{\:\sqrt{\mathrm{4}{x}}}}\:=\mathrm{1} \\ $$

Answered by puissant last updated on 21/Oct/21

1)  L=lim_(n→∞) (((n!)/n^n ))^(1/n) ; ln(L)=lim_(n→∞) (1/n)ln(((n!)/n^n ))  = lim_(n→∞) (1/n){ln(n!)−nln(n)}  = lim_(n→∞) (1/n)Σ_(k=1) ^n {ln(k)−ln(n)}  =lim_(n→∞) (1/n)Σ_(k=1) ^n ln((k/n)) = ∫_0 ^1 ln(x)dx  IBP →  { ((u=lnx)),((v′=1)) :} ⇒  { ((u′=(1/x))),((v=x)) :}  ⇒ ln(L)=[xlnx]_0 ^1 −∫_0 ^1 1dx  ⇒ ln(L)=−1 → L=e^(−1) =(1/e)..        ∴∵ L=lim_(n→∞) (((n!)/n^n ))^(1/n) = (1/e)..                ...........Le puissant............

$$\left.\mathrm{1}\right) \\ $$$$\mathscr{L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} ;\:{ln}\left(\mathscr{L}\right)=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}{ln}\left(\frac{{n}!}{{n}^{{n}} }\right) \\ $$$$=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\left\{{ln}\left({n}!\right)−{nln}\left({n}\right)\right\} \\ $$$$=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left\{{ln}\left({k}\right)−{ln}\left({n}\right)\right\} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\frac{{k}}{{n}}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({x}\right){dx} \\ $$$${IBP}\:\rightarrow\:\begin{cases}{{u}={lnx}}\\{{v}'=\mathrm{1}}\end{cases}\:\Rightarrow\:\begin{cases}{{u}'=\frac{\mathrm{1}}{{x}}}\\{{v}={x}}\end{cases} \\ $$$$\Rightarrow\:{ln}\left(\mathscr{L}\right)=\left[{xlnx}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{1}{dx} \\ $$$$\Rightarrow\:{ln}\left(\mathscr{L}\right)=−\mathrm{1}\:\rightarrow\:\mathscr{L}={e}^{−\mathrm{1}} =\frac{\mathrm{1}}{{e}}.. \\ $$$$\:\:\:\:\:\:\therefore\because\:\mathscr{L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} =\:\frac{\mathrm{1}}{{e}}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:...........\mathscr{L}{e}\:{puissant}............ \\ $$

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