Question Number 205174 by universe last updated on 12/Mar/24
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left\{{x}^{\mathrm{1}/{x}} \right\}\:=\:?\:{where}\:\left\{.\right\}\:{is}\:{a}\:{fractional}\:{part}\:{of}\:{x} \\ $$
Answered by lepuissantcedricjunior last updated on 12/Mar/24
$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\boldsymbol{{x}}\right)^{\frac{\mathrm{1}}{\boldsymbol{{x}}}} =\infty^{\mathrm{0}} =\boldsymbol{\mathrm{F}{I}} \\ $$$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\left(\boldsymbol{{x}}\right)^{\frac{\mathrm{1}}{\boldsymbol{{x}}}} =\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\boldsymbol{{e}}^{\frac{\boldsymbol{{lnx}}}{\boldsymbol{{x}}}} =\boldsymbol{{e}}^{\mathrm{0}} =\mathrm{1} \\ $$$$\boldsymbol{{car}}\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{\boldsymbol{{lnx}}}{\boldsymbol{{x}}}=\mathrm{0}\:\:\boldsymbol{{formule}}\:\boldsymbol{{usuelle}} \\ $$$$………{le}\:{puissant}\:\boldsymbol{{D}}{r}………… \\ $$
Answered by Berbere last updated on 13/Mar/24
![=x^(1/x) −[x^(1/x) ] x→^f x^(1/x) ; f(x)=e^((1/x)ln(x)) ;lim_(x→∞) f(x)=1 x>1⇒f(x)≥1 ⇒∃A∈R,∀x≥A⇒1≤f(x)≤(3/2) ⇒∀x≥A [e^((ln(x))/x) ]=1 ⇒x^(1/x) −[x^(1/x) ]=x^(1/x) −1;∀x≥A lim_(x→∞) x^(1/x) −1=0](https://www.tinkutara.com/question/Q205195.png)
$$={x}^{\frac{\mathrm{1}}{{x}}} −\left[{x}^{\frac{\mathrm{1}}{{x}}} \right] \\ $$$${x}\overset{{f}} {\rightarrow}{x}^{\frac{\mathrm{1}}{{x}}} ; \\ $$$${f}\left({x}\right)={e}^{\frac{\mathrm{1}}{{x}}{ln}\left({x}\right)} ;\underset{{x}\rightarrow\infty} {\mathrm{lim}}{f}\left({x}\right)=\mathrm{1} \\ $$$${x}>\mathrm{1}\Rightarrow{f}\left({x}\right)\geqslant\mathrm{1}\:\Rightarrow\exists{A}\in\mathbb{R},\forall{x}\geqslant{A}\Rightarrow\mathrm{1}\leqslant{f}\left({x}\right)\leqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\forall{x}\geqslant{A}\:\:\left[{e}^{\frac{{ln}\left({x}\right)}{{x}}} \right]=\mathrm{1} \\ $$$$\Rightarrow{x}^{\frac{\mathrm{1}}{{x}}} −\left[{x}^{\frac{\mathrm{1}}{{x}}} \right]={x}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1};\forall{x}\geqslant{A} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{x}^{\frac{\mathrm{1}}{{x}}} −\mathrm{1}=\mathrm{0} \\ $$