Neglecting Terms Based On Their Relevant Magnitude If we have an expression like:
$$\frac{1}{l}[ \frac{1}{a} - \frac{1}{a+l}]$$ 
and we know that $l<<a$ then we can convert the above expression to
$$ \frac{1}{a(a+l)}$$ 
and then neglect the l from the denominator. So we end up with:
$$ \frac{1}{a^2} $$
But this is a very easy case. A more complicated case whould be this:
$$ \frac{1}{ \sqrt{(β-α-l)^2+γ^2}}-\frac{1}{ \sqrt{(β-α)^2+γ^2}} $$
Where $ l<<β-l-a$
My question is what rules do we follow to make those neglections?
I assume these hold:


*

*$a+l \approx a$

*$\frac{a}{l} \to \infty$

*$\frac{l}{a} \approx 0$
Present examples if possible.
 A: Hint. One may write, by the Taylor series expansion, as $\dfrac{l}{a} \to 0$,
$$
\frac{1}{l}\left[ \frac{1}{a} - \frac{1}{a+l}\right]=\frac{1}{a(a+l)}=\frac1{a^2}\cdot\frac{1}{1+\dfrac{l}{a}}=\frac1{a^2}\left[1-\frac{l}{a}+O\left(\frac{l^2}{a^2}\right)\right]=\frac1{a^2}+O\left(\frac{l}{a^3}\right). \tag1
$$ Similarly, as $\dfrac{l}{β-α} \to 0$, one has
$$
\begin{align}
& \frac{1}{ \sqrt{(β-α-l)^2+γ^2}}-\frac{1}{ \sqrt{(β-α)^2+γ^2}}
\\\\&=\left(\frac{1}{ \sqrt{(β-α)^2+γ^2-2l\cdot(β-α)+l^2}}-\frac{1}{ \sqrt{(β-α)^2+γ^2}}\right)
\\\\&=\frac{1}{\sqrt{(β-α)^2+γ^2}}\left[\frac{1}{\sqrt{1-2\dfrac{l\cdot(β-α)}{(β-α)^2+γ^2}+\dfrac{l^2}{(β-α)^2+γ^2}}}-1\right]
\\\\&=\frac{1}{\sqrt{(β-α)^2+γ^2}}\left[1+\dfrac{l\cdot(β-α)}{(β-α)^2+γ^2}+O\left(\dfrac{l^2}{(β-α)^2}\right)-1\right]
\\\\&=\frac{1}{\sqrt{(β-α)^2+γ^2}}\cdot O\left(\dfrac{l}{β-α}\right)
\\\\&=O\left(\dfrac{l}{β-α}\right) \tag2
\end{align}
$$ where we have used the standard Taylor series expansion, as $x \to 0$,
$$
\frac1{\sqrt{1-x}}=1+\frac{x}2+O(x^2).
$$
