I’m going over Griffiths EM example 3.9 (different question on this example is here Metal Sphere in a Uniform Electric field).
I’m wondering where the logic for $$V\rightarrow -E_0z= -E_0r\cos(\theta) \quad \text{ for } r\gg R$$ is coming from. Shouldn’t this potential behave like a dipole at very large $r$, i.e $$V(r)\approx \frac{1}{4\pi\epsilon_0}\frac{qR\cos(\theta)}{r^2} \quad ?$$
You are correct in that the dipole field is present but the constant field dominates.
$$ \lim_{r \to \infty}r = \infty $$
$$ \lim_{r \to \infty}\frac{1}{r^2} = 0 \quad \text{(or tends to zero)} $$
When matching conditions for large $r$ we need to match the dominant term.
In a mode expansion approach to the potential (which seems to be what you are looking at) we include all terms with same mode order, so the dipole term is present in the final answer and is needed to make the potential constant on the surface of the conducting sphere.
If you had some other type of mode problem without a uniform external field you may need to assume a monopole or dipole behavior at infinity to justify keeping or eliminating various terms in the expansion.
The $V(r,\theta)= -E_{\infty} r \cos \theta$ part is just the potential of the uniform background field. The dipole field has decayed away at large distance.
