Accurate Formula and One Old-Exam Questions?! We get stuck in a problem on old-exam. 
\begin{equation*}
A=\sqrt{x+ \frac {2}{x}} -\sqrt{x- \frac {2}{x}}~\text{and}~x>>1. 
\end{equation*}
For calculating $A$ which of the following option give a more accurate for primary formula ? 
$ I) \frac{1}{x\sqrt{x}}$ for $x>\sqrt{\frac{2}{u}}$
$ II) \frac{2}{x\sqrt{x}}$ for $x>\sqrt{\frac{2}{u}}$
$ III) \frac{1}{x\sqrt{x}}$ for $x>\frac{2}{u}$
$ IV) \frac{2}{x\sqrt{x}}$ for $x>\frac{1}{u}$

The important aspect of this question is that $U$ not defined. how can
  describe it and say which of the option is true?

 A: The numerically stable formula for that expression is obtained via binomial formulas, 
$$
A=\sqrt{x+ \frac {2}{x}} -\sqrt{x- \frac {2}{x}}
=\frac{\left(x+ \frac {2}{x}\right) -\left(x- \frac {2}{x}\right)}{\sqrt{x+ \frac {2}{x}} + \sqrt{x- \frac {2}{x}}}
\\
=\frac{2}{x\sqrt{x}}·\frac{2}{\sqrt{1+ \frac {2}{x^2}} + \sqrt{1- \frac {2}{x^2}}}
$$
Now $$\sqrt{1+h}=1+\frac12 h-\frac18h^2+\frac1{16}h^3-\frac{5}{128}h^4+…$$ so that $$\sqrt{1+h}+\sqrt{1-h}=2\left(1-\frac18h^2-\frac{5}{128}h^4-…\right)$$ and
$$
A=\frac{2}{x\sqrt{x}}·\frac{1}{1-\frac18·\frac4{x^4}+O(x^{-8})}
$$
Thus whenever $\frac1{2x^4}<\frac{\mu}2$ or $x>\sqrt[4\,]{\frac1\mu}$ the value of $A$ is accurately represented by the first factor, which gives an asymptotically correct formula.

Here $\mu$ is the smallest positive number so that in the floating point type $1+\mu$ is different from $1$ is different from $1-\mu/2$. For the double type, $\mu=2^{-52}\simeq 2·10^{-16}$ is a little larger than $10^{-16}$, which means that already for $x>10^{4}$ the short formula is correct within the floating point precision.

Diagram comparing the absolute numerical and approximation errors:

and the relative errors (showing that $x>10^4$ is a reasonable bound) (note that the errors of the original expression had to be scaled down radically to even be visible)

where $f(x)$ is the original expression for $A$, $g(x)$ the stabilized expression per the binomial formula and $h(x)$ the approximation by the first factor.

Another approach
The original motivation for this selection of possibilities may be that for $x>\sqrt{\frac2\mu}$ the term $\frac2x$ is so small that it does not contribute any digits to the expressions $x\pm \frac2x$, i.e., $fl(x\pm \frac2x)=fl(x)$, where $fl$ denotes rounding to floating point precision. At that point $$fl(\sqrt{fl(x+2/x)})-fl(\sqrt{fl(x-2/x)})=0,$$ which obviously is not acceptable as numerical result. 
So to get any kind of reasonable result, one needs a modified formula for larger $x$. The additional question one should ask is if an earlier use of the asymptotic expression would yield better precision than the original formula? To which the (empiric) answer is: Yes, already for $x>1000$ (surely, $x>500$  mostly) the asymptotic formula has a better error than the original formula. 

Theoretically, the relative floating point error of the original formula is $\sim \mu x^2$, the relative error of the asymptotic formula $\sim x^{-4}$, thus the cross-over point is at $\sqrt[6\,]{\frac1\mu}\simeq 4.7·10^2$, which coincides with the empirical results.
A: I) and III) can be dropped immediately for being "wrong".
The error in using $\sqrt{x\pm\frac 2x+\frac1{x^3}}=\sqrt x\pm\frac1{x\sqrt x}$ is fairly small, namely approximately $\frac1{x^2}$ times the derivative of the square root at $x\pm\frac2x\approx x$, i.e., $\approx \frac1{2x^2\sqrt x}$. Hence the total error from both summands is $\approx \frac1{x^2\sqrt x}$, and in fact the errors are more cancelling than adding up. Hence if an error of $u$ is acceptable, $x>\sqrt{\frac1u}$ is certainly sufficient.
