Difference between Taylor polynomial of $f(r)=(1+\frac{a^2-2ay}{r^2})^{-3/2}$ at $r=m$ as $m\to \infty$ and the function $(1+s)^{-3/2}$ at $s=0$?

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Consider the expression

$$E(x,y)=\frac{x\hat{i}+(y-a)\hat{j}}{(x^2+(y-a)^2)^{3/2}}\tag{1}$$

where $r^2=x^2+y^2$. I'd like to investigate what happens when $r>>a$.

Let's consider the factor involving just the denominator

$$(x^2+(y-a)^2)^{-3/2}$$

$$=(x^2+y^2-2ay+a^2)^{-3/2}$$

$$=\left [ (x^2+y^2)(1+\frac{a^2-2ay}{x^2+y^2})\right ]^{-3/2}$$

$$=r^{-3}\left ( 1+ \frac{a^2-2ay}{r^2} \right )^{-3/2}\tag{2}$$

Let $s(r)=\frac{a^2-2ay}{r^2}$ and let $f(x)=(1+x)^{-3/2}$.

Then $f(s(r))=\left ( 1+ \frac{a^2-2ay}{r^2} \right )^{-3/2}\tag{3}$

Since $s\to 0$ as $r\to \infty$, apparently we can use a Taylor approximation to $f(x)$ near $0$ to find out what happens to (3) as $r\to\infty$.

$$P_{1,f,0}=1-\frac{3}{2}s\tag{4}$$

This is a linear approximation to $f$ near $x=0$. When $r\to\infty$ and $s\to 0$, $f(s(r))\approx 1-\frac{3}{2}s$.

Which means that when $r\to\infty$ the expression in (2) is approximately

$$r^{-3}(1-\frac{3}{2}s)$$

$$=r^{-3}\left (1-\frac{3}{2}\frac{a^2-2ay}{r^2}\right )\tag{5}$$

which we can sub into (1) to reach the desired expression for $E$ when $r>>a$.

What is the theorem or theorems that justify this type of calculation when we have a function composition?

What do I have to study to learn more about this specific topic of using Taylor approximations like they do in physics?

I recall from studying Taylor polynomials and Taylor's Theorem a formula for computing the Taylor polynomial of a function composition. The expression in (3) is a function composition.

If we apply the formula

$$P_{n,a,f\circ s}=\left [ P_{n,s(m),f}\circ P_{n,m,s} \right ]_n$$

where the $[]$ operator means truncation of terms that are of degree $\leq n$, then we reach the second-to-last result seen in the following Maple session (which includes all the calculations mentioned above):

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Equation (12) above is the first-order Taylor polynomial of the function composition in (3). But it seems like this result isn't as useful as the previous one obtained in (5).

The variable $m$ in (12) represents the value about which we computed the Taylor polynomial of the function composition.

Is there some useful way to relate this calculation with the one that led to (5)?