Deducing the approximation using double greater-than

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I've been struggling against the below deduction of the approximation.

$0<a\in \mathbb{R}$

$d:=\sqrt{x^2+y^2} \hspace{0.3cm}$ where $ \hspace{0.1cm} 0\leq x,y\in \mathbb{R}$

$a\ll d$

$$\frac{1}{\sqrt{1-\frac{2ax}{d^2}}}+\frac{1}{\sqrt{1+\frac{2ax}{d^2}}}\simeq \left(\left( 1+\frac{ax}{d^2}\right)-\left(1-\frac{ax}{d^2}\right)\right)$$

Can anyone tell me some hint(s) so that I can deduce it myself?

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Hint:

Can you show $\frac{2ax}{d^2} \ll 1$?

If so, then since $(1+\epsilon)^{-1/2} = 1 -\frac{\epsilon}{2} + O(\epsilon^3)$, you can compute

$$ \frac{1}{\sqrt{1+\frac{2ax}{d^2}}} = \left ( 1 + \frac{2ax}{d^2} \right )^{-1/2} = 1 - \frac{ax}{d^2} + O \left ( \left ( \frac{2ax}{d^2} \right )^3 \right ) $$

And similarly for $\left ( 1 - \frac{2ax}{d^2} \right )^{-1/2}$.

Do you see how to finish the problem from here?


I hope this helps ^_^