The Bessel DE: $$z^2\frac{\text d^2f}{\text{d}z^2}+z\frac{\text{d}f}{\text{d}z}+\left(z^2-m^2\right)f = 0.$$ The Bessel DE can be rewritten as:
$$\frac{d^2f}{\text{dz}^2} + a(z)\frac{df}{ dz } + b(z)f=0.$$
At the neighbourhood of $z_0$, we make the assumption that
$$z^2f \ll m^2f,$$
even if $m =0$.
Can someone explain why? If $m=0$ no matter how small $z$ is, it does seem bigger than $m^2f$.
$m^2f$ would be zero if $m=0$, all the assumption requires is that $f$ becomes negative somewhere, which it does: https://www.wolframalpha.com/input/?i=J_0
I guess it depends on what you mean by $"<<"$, and also I assume you just want there to exist some $z_0$ such that this holds (because there is no $z_0$ in your equation).