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For $x \in (0,1]$, Since $$\lim\limits_{x\to0} \frac{\Big|\ln\frac{1}{x}\Big|^k}{\frac{1}{x}}=0 \text{ for }k>1$$

Why is it true that $\exists x\in (0,b]$ such that $\Big|\ln\frac{1}{x}\Big|^k \leq \frac{1}{x}$ and $0<b<1$?

Web Typer
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  • See for example https://math.stackexchange.com/q/1138996/42969 – Martin R Mar 28 '23 at 12:12
  • @MartinR Why is it true that $\exists x\in (0,b]$ such that $\Big|\ln\frac{1}{x}\Big|^k \leq \frac{1}{x}$ and $0<b<1$? – Web Typer Mar 28 '23 at 12:15
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    Something is missing in the first equation. You probably mean $\lim\limits_{x\to0} \frac{\Big|\ln\frac{1}{x}\Big|^k}{\frac{1}{x}} = 0$. Which means that $\frac{\Big|\ln\frac{1}{x}\Big|^k}{\frac{1}{x}} < 1$ for sufficiently small $x$ ... – Martin R Mar 28 '23 at 12:16
  • @MartinR exactly. – Web Typer Mar 28 '23 at 12:22

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