<s>[INST] 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?
 [/INST] 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:
$$L=\lim_{x\rightarrow \infty} \frac{1}{\sqrt{\frac{1+\frac{\sqrt{x+\sqrt{x}}}{x}}{1+\frac{\sqrt{x}}{x}}}+\sqrt{\frac{1}{1+\frac{1}{\sqrt{x}}}}}=\frac{1}{\sqrt{\frac{1+0}{1+0}}+\sqrt{\frac{1}{1+0}}}=\frac{1}{2}$$
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