difference of square roots approximation In two of my physics courses in the past week, I've come across an approximation for the difference of two square roots for large radicands:
$\sqrt{x+a}-\sqrt{x+b}\approx\frac{a-b}{2\sqrt x}$ for large $x$.
I have no idea where this comes from.
Wolfram|Alpha gives me the simplest answer by handing me this as the first term of the expansion around $x=\infty$, but how is that calculated? I know how normal Taylor Series expansions are formulated, but I don't get how to run this calculation.
Just trying to get the approximation alone, the closest I've gotten is with the logic:
$$
\frac{\sqrt{x+b+(a-b)}-\sqrt{x+b}}{a-b}\approx\frac{d}{dx}\sqrt{x+b}
\\
\sqrt{x+a}-\sqrt{x+b}\approx\frac{a-b}{2\sqrt{x+b}}
$$
But, as you can tell, the radicand in the approximation is incorrect. (Not to mention the approximation there is that $a\approx b$, not large $x$.) I can try to say that for large $x$, $\sqrt{x+b}\approx\sqrt{x}$, but that doesn't explain why the version with just $x$ as the radicand is more accurate. (Which is true.) Plus, I also want to know how to get the higher-order terms in the expansion.

Factor out $x$ to get
$$\sqrt{x}\left(\sqrt{1 + \frac{a}{x}} - \sqrt{1 + \frac{b}{x}}\right)$$
Now expand each square root in a Binomial series
$$=\sqrt{x}\left(1 + \frac{a}{2x} - 1 - \frac{b}{2x} + \mathcal{O}(x^{-2})\right)$$
For large $x$ we can neglect the terms of order $x^{-2}$ and smaller to make the approximation
$$\sqrt{x + a} - \sqrt{x + b} \approx \frac{a-b}{2\sqrt{x}}$$
