Landau symbol and taylor series $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=\frac{1}{2}$$
I saw the calculation $$\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=$$...$$=\sqrt x (\sqrt{1+\sqrt{1/x+\sqrt{1/x^3}}}-1)= $$ $$\sqrt x(1+\frac{1}{2}\sqrt{1/x+\sqrt{1/x^3}}+ \color{red}{\mathcal o \left(1/x+\sqrt{1/x^3}\right)}-1)=...$$
Main question: Now wouldn't it have to be $\mathcal o(\sqrt{1/x+\sqrt{1/x^3}})$ instead?  So I could write $\mathcal o(\sqrt{1/x+\sqrt{1/x^3}})$ or $\mathcal O(1/x+\sqrt{1/x^3})$?
Are there other ways to calculate the limit?

An overkill is to exploit that for any sufficiently large $x>0$ we have that
$$ \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots}}}} = \frac{1}{2}+\sqrt{x+\frac{1}{4}}$$
since both terms are a positive solution of $y^2=x+y$. On the other hand
$$ \sqrt{x+\sqrt{x}} \geq \frac{1}{2}+\sqrt{x-\frac{1}{4}} $$
is a trivial inequality by just squaring both sides, hence
$$ \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}-\frac{1}{2} $$
is bounded between $\sqrt{x-\frac{1}{4}}-\sqrt{x}$ and $\sqrt{x+\frac{1}{4}}-\sqrt{x}$, and the limit is $\color{red}{\large\frac{1}{2}}$ by squeezing.
