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?

The Taylor series is
$$ \sqrt{a+x} - \sqrt{a} = \sum_{k=1}^\infty (-1)^{k+1} \frac{(2k)!}{(k!)^2(2k-1)} 4^{-k} a^{1/2-k} x^k$$
If $|x/a| < 1$, the absolute values of the terms decrease, since if $c_k = (2k)!/((k!)^2 (2k-1) 4^k)$, 
$$ \frac{c_{k+1}}{c_k} = \frac{2k-1}{2k+2} < 1$$
Thus if $a > x > 0$ the absolute value of the error is always less than that of the next term.
However, if $x/a$ is close to $1$ the convergence is rather slow:
$$c_k \sim \frac{1}{2 \sqrt{\pi} k^{3/2}}$$
so that won't be less than $2^{-53}$ unless $k > 1.862 \times 10^{10}$ approximately.
