Polynomial bound for the log function

Question

For $x>0$, is $\log{x}$ bounded by $x^{-\epsilon}$ for any $\epsilon>0$ when x is small?

Answer 1

Yes,because $\frac {\log x}{x^{-\epsilon}} \to 0$ by L'Hopital's Rule.

Answer 2

Let's try to show that $\log x = o(\varepsilon)$ for suitable $\varepsilon$.

Let us first prove the probably more common fact, any growing exponential function grows faster than a monomial.

For any $\alpha>0$, $\beta > 1$, we can choose an integer $c\gt \alpha$. Applying L'Hôpital's rule $c$ times we get

$$0\le\lim_{x\to\infty}\frac{x^\alpha}{\beta^x}\le\lim_{x\to\infty}\frac{x^c}{\beta^x} 0=\lim_{x\to\infty}\frac{cx^{c-1}}{(\log\beta)\beta^x}\\ =\lim_{x\to\infty}\frac{cx^{c-1}}{(\log\beta)\beta^x}=\lim_{x\to\infty}\frac{c(c-1)x^{c-2}}{(\log\beta)^2\beta^x}\\ =\cdots=\lim_{x\to\infty}\frac{c(c-1)\cdots2\cdot1}{(\log\beta)^c\beta^x}=0$$ where the last equality holds since $\beta^x = (1+(1-\beta))^x\gt1+x(1-\beta)\to\infty$ as $x\to\infty$.

So, $x^\alpha = o(\beta^x)$ for any $\alpha>0$, $\beta > 1$.

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Let $\alpha = 1, \beta=e^\varepsilon$. By a change of variable $x=\log n$, we can see that $$\log n = o(e^{\varepsilon\log n})=o(n^\varepsilon)$$ Similarly, we have $$\log\log n = o((\log n)^\varepsilon) \quad\text{ for any }\varepsilon\gt 0 $$ $$e^n = o(\lambda^{e^n}) \quad\text{ for any }\lambda\gt 1 $$