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?

It was pointed out by Robert Israel that the series does badly when $|x| \approx |a|$, but in that case the loss of significance of the naive evaluation is small.
It was also suggested by Winther (and a since-deleted answer) to rewrite as $$\frac{x}{\sqrt{a+x}+\sqrt{a}}$$
The series for the denominator is similar to the series in the question, only with a leading constant term.  This means that when $\left|\frac{x}{a}\right|$ is small enough, the addition of terms eventually becomes insignificant in double arithmetic.
If $\left|\frac{x}{a}\right| < 2^{-52}$, $1$ term is sufficient.  Otherwise
If $\left|\frac{x}{a}\right| < 2^{-25}$, $2$ terms are sufficient.  Otherwise
If $\left|\frac{x}{a}\right| < 2^{-16}$, $3$ terms are sufficient.  Otherwise
If $\left|\frac{x}{a}\right| < 2^{-11}$, $4$ terms are sufficient.  Otherwise
If $\left|\frac{x}{a}\right| < 2^{-9}$, $5$ terms are sufficient.  Otherwise
If $\left|\frac{x}{a}\right| > 2^{-9}$, the loss of precision in the addition $a + x$ is relatively small.
But in fact, perhaps $$\frac{x}{\sqrt{a + x} + \sqrt{a}}$$ evaluated in double precision is good enough for all $|x| << |a|$ and series are unnecessary?
