<s>[INST] How many Taylor series terms are needed to accurately approximate $\sqrt{a+x}-\sqrt{a}$? Naive evaluation of $\sqrt{a + x} - \sqrt{a}$ when $|a| >> |x|$ suffers from catastrophic cancellation and loss of significance.
WolframAlpha gives the Taylor series for $\sqrt{a+x}-\sqrt{a}$ as:
$$\frac{x}{2 \sqrt{a}} - \frac{x^2}{8 a^{3/2}} + \frac{x^3}{16 a^{5/2}} - \frac{5 x^4}{128 a^{7/2}} + \frac{7 x^5}{256 a^{9/2}} + O(x^6)$$
which (I think) equals:
$$\sqrt{a} \left( \frac{1}{2} \left(\frac{x}{a}\right) - \frac{1}{8} \left(\frac{x}{a}\right)^2 + \frac{1}{16} \left(\frac{x}{a}\right)^3 - \frac{5}{128} \left(\frac{x}{a}\right)^4 + \frac{7}{256} \left(\frac{x}{a}\right)^5 + O\left(\left(\frac{x}{a}\right)^6\right) \right)$$
How quickly do the coeffients decrease?
How many terms are needed to reach $53$ bits of accuracy (IEEE double precision) in the result given that $10^{-300} < \left|\frac{x}{a}\right| < 1$ is known?
Alternatively, what are the threshold values of $\left|\frac{x}{a}\right|$ where the number of terms changes?
What about rounding errors, assuming each value is stored in double precision?
 [/INST] To avoid cancellation error the first thing to do is to write:
$$
\sqrt{a+x}-\sqrt{a}=\frac{x}{\sqrt{a+x}+\sqrt{a}}=\sqrt{a}\frac{x}{a}
\frac{1}{1+\sqrt{1+\frac{x}{a}}}
$$
then with $y=\frac{x}{a}$ you must approximate this
$$
\sqrt{a}\frac{y}{1+\sqrt{1+y}}
$$ 
fonction for $y\in[10^{-300},1]$. This function has nothing pathological and IMHO can be computed in a straightforward way.
If you really want to use Taylor series for $y\sim 0$
$$
\sqrt{a}\frac{y}{1+\sqrt{1+y}}=\sqrt{a}(\frac{y}{2}-\frac{y^2}{8}+\frac{y^3}{16}-\frac{5 y^4}{128}+\frac{7
   y^5}{256}+O\left(y^6\right))
$$
I assume that the series is alternating, hence the error term $e$ is majored by $|e|<\sqrt{a}\frac{7y^5}{256}$. For instance if you want $|e|<10^{-q}$ you can use the Taylors series for $0\le y \le y_*$ where $y_*$ is such that 
$$\sqrt{a}\frac{7y_*^5}{256}<10^{-q}$$ which gives
$$y_*<10^{-q/5}(\frac{256}{7\sqrt{a}})^{1/5}$$
Example: with $q=5$, $a=3$
We get $y_*<0.184042$. 
That means that you can use the Taylors series for $ y_*=\frac{x_*}{a}<0.184042$, hence $x_*<3\times 0.184042 \approx 0.552125$.
Let's try with $x=0.55$. 
With the initial formula we find:
$$
\sqrt{a+x}-\sqrt{a}\approx 0.152094
$$
With the Taylor series, with $y=\frac{0.55}{3}$ we get 
$$
\sqrt{a}(\frac{y}{2}-\frac{y^2}{8}+\frac{y^3}{16}-\frac{5 y^4}{128}+\frac{7
   y^5}{256})\approx 0.152095
$$
We see that the error $|e|=|0.152094-0.152095|\approx 1.17957\times 10^{-6}$ is less that $10^{-q}=10^{-5}$ as expected
</s>