I am having troubles showing the above step regarding the approximation $$ \sqrt{\frac{z}{z-2h}} - \sqrt{\frac{z}{z+2h}} \approx \frac{2h}{z}, \quad h>0, |z| \gg h$$ given in an old exam.
My attempt is as follows, Denote $f(z) = \sqrt{\frac{z}{z-2h}} = \frac{1}{\sqrt{1-2h/z}}$, by letting $t=h/z$ we try to approximate $f(t(z))$ for $ t \approx 0$ by a Maclaurin series. We get that,
$$f(t(z)) = \frac{1}{\sqrt{1-2t}} \approx 1 + \frac{t'}{(1-2t)^{3/2}}t$$ where the last term can be written
$$\frac{-h^2}{z^3}\frac{1}{(1-2h/z)^{3/2}}.$$ This is where I get stuck, the suggested solution seems to state that the last term should evaluate to $h/z$. Anyone got any suggestions on how to solve this problem?
I will paste the answer given in the comments so that I can close the question,
with the comment given by Daniel Fisher we expand both terms to, with $ t = h/z$
$$ \frac{1}{\sqrt{1-2t}} - \frac{1}{\sqrt{1+2t}} = (1 + t + O(t^2)) - (1 - t + O(t^2)) \approx 2t = 2 \frac{h}{z}$$ for $|z| \gg h$.