Was getting back into physics and reading a chapter on electrostatics which sets up the following situation. We have a configuration of point charges - one $-q$ at the point ($-d,0,0$) and one $+q$ at the point ($d,0,0$). The potential of this configuration is then just $$V(x) = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{\lvert x-d \rvert} - \frac{q}{\lvert x + d\rvert}\right)$$ Now the book then says "Taylor expanding this expression for $x\gt\gt d$ tells you approximately how the field behaves far away". It's my understanding that performing a Taylor expansion at a point gives one the ability to approximate how a function behaves at that point when simply calculating the value might be too cumbersome or difficult. However, it's been a while since I've actually done a Taylor Expansion and was looking for some help in doing this for myself.
So, I know the general formula for a Taylor Expansion is the following $$f(x)\vert_a = \sum \frac{f^{(n)}(a)}{n!}(x-a)^n$$ Here in this case I'm a bit confused as to how to approach this given the information that we are expanding about a point $x$ such that $x\gt\gt d$ I get that it means we are investigating how the function behaves far away from the point $(d,0,0$) but not sure how this condition meaningfully changes out function. It's my first instinct to say that since $x \gt\gt d$ that adding or subtracting $d$ from $x$ doesn't really make a difference and thus $x + d = x - d =x$ but then that leaves us with $$V(x) = \frac{1}{4\pi\epsilon_0}\left(\frac{q}{x} - \frac{q}{x}\right) = 0$$ Which I know isn't what the example is prompting.
TL;DR
I want to know what the condition $x \gt\gt d$ meaningfully changes about how we proceed with the Taylor expansion of the potential function and then would also like some guidance in carrying out said expansion.