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Question Number 205174 by universe last updated on 12/Mar/24

$$\underset{x\to \mathrm{\infty }}{\lim }\left\{x^{1/x}\right\}=?where\{.\}isafractionalpartofx$$

Answered by lepuissantcedricjunior last updated on 12/Mar/24

$$\underset{x\to +\mathrm{\infty }}{\lim }{\left(\mathit{x}\right)}^{\frac{1}{\mathit{x}}}={\mathrm{\infty }}^0=\mathbf{F}\mathit{I}$$$$\underset{x\to +\mathrm{\infty }}{\lim }{\left(\mathit{x}\right)}^{\frac{1}{\mathit{x}}}=\underset{x\to +\mathrm{\infty }}{\lim }{\mathit{e}}^{\frac{\mathit{l}\mathit{n}\mathit{x}}{\mathit{x}}}={\mathit{e}}^0=1$$$$\mathit{c}\mathit{a}\mathit{r}\underset{x\to +\mathrm{\infty }}{\lim }\frac{\mathit{l}\mathit{n}\mathit{x}}{\mathit{x}}=0\mathit{f}\mathit{o}\mathit{r}\mathit{m}\mathit{u}\mathit{l}\mathit{e}\mathit{u}\mathit{s}\mathit{u}\mathit{e}\mathit{l}\mathit{l}\mathit{e}$$$$.........lepuissant\mathit{D}r............$$

Answered by Berbere last updated on 13/Mar/24

$$=x^{\frac{1}{x}}-\left[x^{\frac{1}{x}}\right]$$$$x\stackrel{f}{\to }{x}^{\frac{1}{x}};$$$$f\left(x\right)=e^{\frac{1}{x}ln\left(x\right)};\underset{x\to \mathrm{\infty }}{\lim }f\left(x\right)=1$$$$x>1\Rightarrow f\left(x\right)⩾1\Rightarrow \mathrm{\exists }A\in \mathbb{R},\mathrm{\forall }x⩾A\Rightarrow 1⩽f\left(x\right)⩽\frac{3}{2}$$$$\Rightarrow \mathrm{\forall }x⩾A\left[e^{\frac{ln\left(x\right)}{x}}\right]=1$$$$\Rightarrow x^{\frac{1}{x}}-\left[x^{\frac{1}{x}}\right]=x^{\frac{1}{x}}-1;\mathrm{\forall }x⩾A$$$$\underset{x\to \mathrm{\infty }}{\lim }{x}^{\frac{1}{x}}-1=0$$

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