Can $O(\sqrt{x})$ be considered $o(x)$?

Question

This example challenges my understanding of $O(x)$ and $o(x)$ notation. One the one hand I have:

$$ A = B + o(x)$$

Another part of the paper uses big-O instead of little-o and says:

$$ C = D + O(\sqrt{x}) \stackrel{?}{=} D + o(x)$$

I am willing to take a huge sacrifice on the error term for simplicity, but I am struggling to see if this is correct. In particular, is it the case that: $\bbox[2px, border:2px solid #55FF88]{ \tfrac{1}{x}(C-D) \asymp 0 }$ or $\bbox[2px, border:2px solid #5588FF]{\tfrac{1}{x}(C-D) \sim 0}$ ?

Answer 1

Yes. If we have $f(x) \in O(\sqrt{x})$, then $f(x) < k\sqrt{x}$ for some $k$ and all $x$ sufficiently large.

Then we have $f(x) < \frac{k}{\sqrt{x}}x$, but since $\frac{k}{\sqrt{x}}$ gets abritarily small, we can conclude that $f(x) < cx$ or all $c$ and all $x$ sufficiently large. This is equivalent to $f(x) \in o(x)$.

We can make much stronger claims. For example $f(x) \in o(\sqrt{x}\ln(x))$ would be true too.

Answer 2

Yes: if $f(x)=O(\sqrt x)$ there exists a constant $K>0$ such that $\lvert f(x)\rvert\le Kx$ for large $x$. This implies $$0\le\biggl\lvert\frac{f(x)}x\biggr\rvert\le\biggl\lvert\frac{K}{\sqrt x}\biggr\rvert $$ which tends to $0$ as $x$ tends to $+\infty$.

For the second point, if $C-D=o(x)$, then $\;\dfrac{C-D}x=o(1)$ so $\lim_{x\to\infty}(C-D)=0$. But we can't say a function is equivalent to $0$; equivalence of functions makes sense only for non-identically $0$ functions.