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.