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?
 A: 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.
A: If you want to do it with an asymptotic expansion, you have to be cautious with the order of expansion. The scale of comparison will be the powers of $\sqrt x$, and we'll expand $\sqrt{1+\sqrt{\dfrac1x+\sqrt{\dfrac1{x^3}}}}$ at order $2\;$ (i.e. up to a term in $1/x$).


*

*$\sqrt{\dfrac1x+\sqrt{\dfrac1{x^3}}}=\dfrac1{\sqrt x}\biggl(\sqrt{1+\dfrac1{\sqrt{x}}} \biggr)=\dfrac1{\sqrt x}\biggl(1+\dfrac1{2\sqrt{x}}+o\Bigl(\dfrac1{\sqrt x}\Bigr) \biggr)=\dfrac1{\sqrt x}+\dfrac1{2x}+o\Bigl(\dfrac1x\Bigr)$, so that:

*$\sqrt{1+\sqrt{\dfrac1x+\sqrt{\dfrac1{x^3}}}}=\sqrt{1+\dfrac1{\sqrt x}+\dfrac1{2x}+o\Bigl(\dfrac1x\Bigr)}=$


\begin{align*}&=1+\dfrac{1}{2\sqrt x}+\dfrac1{4x}-\dfrac18\Bigl(\dfrac1{\sqrt x}+\dfrac1{2x}\Bigr)^2+o\Bigl(\dfrac1x\Bigr)=1+\dfrac{1}{2\sqrt x}+\dfrac1{4x}-\dfrac1{8x}+o\Bigl(\dfrac1x\Bigr)\\
&=1+\dfrac{1}{2\sqrt x}-\dfrac1{8x}+o\Bigl(\dfrac1x\Bigr)
\end{align*}
Thus,
\begin{align*}\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{\mathstrut  x}&=\sqrt{\mathstrut x} \biggl(\sqrt{1+\sqrt{1/x+\sqrt{1/x^3}}}-1\biggr)=\sqrt{\mathstrut x} \biggl(1+\dfrac{1}{2\sqrt x}-\dfrac1{8x}+o\Bigl(\dfrac1x\Bigr)-1\Biggr)\\&=\frac12-\frac1{8\sqrt x}+o\Bigl(\dfrac1{\sqrt x}\Bigr).
\end{align*}
A: Use formula $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$ in order to get:
$$L=\lim_{x\rightarrow \infty} \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt x=\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}.$$ 
Now divide numerator and denumerator with $\sqrt{x+\sqrt{x}}$ and calculate the limit:
$$L=\lim_{x\rightarrow \infty} \frac{1}{\sqrt{\frac{1+\frac{\sqrt{x+\sqrt{x}}}{x}}{1+\frac{\sqrt{x}}{x}}}+\sqrt{\frac{1}{1+\frac{1}{\sqrt{x}}}}}=\frac{1}{\sqrt{\frac{1+0}{1+0}}+\sqrt{\frac{1}{1+0}}}=\frac{1}{2}$$
A: $$
\begin{align}
\lim_{x\to\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)
&=\lim_{x\to\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\
&=\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\\
&=\lim_{x\to\infty}\frac{\sqrt{1+\sqrt{\frac1x}}}{\sqrt{1+\sqrt{\frac1x+\sqrt{\frac1{x^3}}}}+1}\\
&=\frac12
\end{align}
$$
A: You are correct that $o(1/x + \sqrt{1/x^3})$ is incorrect there. Given the calculation being done, I would have expected either $O(1/x + \sqrt{1/x^3})$ to appear there, or $o(\sqrt{1/x + \sqrt{1/x^3}})$, as you said. It's probably just a typo.
