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?
 A: 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
A: 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.
A: 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?
