Taylor expansion of the electrostatic potential $1/\|\cdot \|$ I have stumbled over this problem several times in electrodynamics, and I just don't get the hang of it.
The task is to do a Taylor expansion of $\,f(\vec{x},\,\vec{a}) = \frac{1}{\|\,\vec{x}-\vec{a}\|}$, for $\|\,\vec{a}\| \ll \|\,\vec{x}\|$. What I do, is
$$ \frac{1}{\left\|\,\vec{x}-\vec{a}\right\|} = \frac{1}{\left\|\,\vec{x}\right\| \left\|\,\hat{x}- \frac{\left\|\,\vec{a}\right\|}{\left\|\,\vec{x}\right\|}
\hat{a}\right\|}$$
In one dimension, I could now say that one number is much smaller than the other, and hence taylor it around zero. But here we're talking vectors, and I don't see which rules apply here. I don't want to write out the norms in terms of square roots, this is really messy, but rather use some rules like $\nabla \frac{1}{\|\,\vec{x}\|} = \frac{\vec{x}}{\|\,\vec{x}\|^3}$.

It is a bit messy. Luckily, you are not the first person who wanted to have a Taylor series for the electrostatic/Newtonian potential. Legendre worked this out in the 18th century. 

But here we're talking vectors, and I don't see which rules apply here.

Yes, it's a bit confusing. This is why one fixes the direction of $\vec a$ (really, just the angle between $\vec a$ and $\vec x$), and then works with the single-variable function whose argument is the magnitude of $\vec a$. See the article linked above. 
