$o\left(x^{1+\varepsilon}\right)$ for all $\varepsilon>0$ implies $O(x)$?

Question

Suppose that I have shown a function $f$ is $o\left(x^{1-\varepsilon}\right)$ for all $\varepsilon>0$. Can I conclude that $f$ is $O(x)$? This seems intuitively right, but I can't seem to furnish a formal proof.

EDIT: The above should read $o\left(x^{1+\varepsilon}\right)$ for all $\varepsilon>0$. (Actually, I have shown $o\left(x^{\varepsilon-1}\right)$ and I want to show $O\left(x^{-1}\right)$ but I assume the exponent doesn't matter. The right sign for $\varepsilon$ clearly does!)

EDIT 2: As $x \rightarrow \infty$.

You can conclude $f(x)=O(x)$, but this is actually weaker than what you already proved. — Clayton, Aug 27 '21 at 00:13
Right you all are. Apologies, I actually meant that I had shown the function was $o(x^{1+\epsilon})$ for all $\epsilon>0$, and I was hoping to show $O(x)$. EDIT: As $x\rightarrow \infty$. — funklejo, Aug 27 '21 at 00:36
What @BrianMoehring said. So far the OP did not specify $x \to 0$ or $x \to\infty$. — GEdgar, Aug 27 '21 at 00:37

Ravi Fernando · Accepted Answer · 2021-08-27T05:21:45.890

5

No. As $x \to \infty$, the function $f(x) = x \log x$ is $o(x^{1+\varepsilon})$ for all $\varepsilon > 0$ by l'Hospital's rule, but it is not $O(x)$ as $\lim_{x \to \infty} \log x = \infty$. Similarly, $g(x) = \frac{\log x}{x}$ is $o(x^{\varepsilon - 1})$ for all $\varepsilon > 0$, but it is not $O(x^{-1})$.

edited Aug 27 '21 at 05:21

answered Aug 27 '21 at 00:48

Ravi Fernando

7,284

1

Good answer. But why not write "$x \to \infty$" in there? – GEdgar Aug 27 '21 at 00:52
@GEdgar fair enough, I've added it. – Ravi Fernando Aug 27 '21 at 05:25

$o\left(x^{1+\varepsilon}\right)$ for all $\varepsilon>0$ implies $O(x)$?

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