Here is the definition of hayman admissibility according to generatingfunctionology (188-189, pdf online):
Let $f(z) = \sum_{n=0}^{\infty} a_n z^n$ be regular in $|z| < R$, where $0 < R \leq \infty$. Suppose the following conditions hold:
(a) There exists an $R_0 < R$ such that $f(r) > 0$ for $R_0 < r < R$.
(b) There exists a function $\delta(r)$ defined for $R_0 < r < R$ such that $0 < \delta(r) < \pi$ for those $r$, and as $r \rightarrow R$ uniformly for $|\theta| \leq \delta(r)$, we have:$$ f(re^{i\theta}) \sim f(r) e^{i\theta a(r) - \frac{1}{2} \theta^2 b(r)} $$
(c) Uniformly for $\delta(r) \leq |\theta| \leq \pi$, we have:$$ f(re^{i\theta}) = \frac{o(f(r))}{\sqrt{b(r)}} \cdot \quad \text{as} \quad r \rightarrow R $$
(d) As $r \rightarrow R$, we have $b(r) \rightarrow +\infty$. Also,
$$a(r) = r \frac{f'(r)}{f(r)} \quad b(r) = ra'(r) $$Then, we say that $f(z)$ is admissible.
In the case of $f(z) = e^z, a(r) = r, b(r) = r$.
$b(r)$ only approaches infinity as $r \to \infty$, so $R = \infty$ because of condition d (no finite value of $R$ blows up). This value of $R$ is admissible(lol) because of the bounds on $R$.
Because of condition b, $\delta(r)$ must be a positive function that tends to 0. This is because condition b now states that
$$e^{r e^{i\theta}} \sim e^{r(1+i\theta - \frac{\theta^2 }{2})} \quad (r\to R = \infty)$$
uniformly for $|\theta| \leq \delta(r)$. The RHS has a truncated taylor expansion of $e^{i\theta}$, meaning $\delta$(r) has to tend to 0.
But if that's so, then condition c states that, uniformly for $\delta(r) \leq |\theta| \leq \pi$,
$$\sqrt{r} e^{r e^{i\theta}} = o(e^r)$$
Taking absolute values of both sides,
$$\sqrt{r} e^{r\cos(\theta)} = o(e^r)$$
But because $\delta(r)$ is a positive function tending to 0, then shouldn't uniformity be broken because the LHS of this asymptopic statement would get arbitrarily close to $e^r$ in a sense? I think I'm having trouble understanding this uniformity condition, and how it relates to $r \to R = \infty$.
I know things should work out because it's known that $e^z$ is Hayman admissible. A walkthrough or explanation or anything would very much be appreciated. Thank you very much.