If we fix $θ_∗ > 0$, then $P(r, θ) → 0$ uniformly on the set $ \left\lbrace θ : |θ| ≥ θ_∗ \right\rbrace $ as $r → a^-$ $$P(r,\theta) = \frac{a^2-r^2}{a^2-2r\cos(\theta)+r^2}$$ $0\leq r <a,\textrm{ } -\pi \leq \theta \leq \pi$
(Hint: $a^2 − 2ar cos(θ) + r^2 = (a − r)^2 + 2ar(1 − \cos θ)$.)
So far I've gotten:
$ \lim_{r → a^-} \max_{|θ| ≥ θ_∗} \left | \frac{a^2-r^2}{(a − r)^2 + 2ar(1 − \cos θ)} \right |$
$ \lim_{r → a^-} \left | \frac{a^2-r^2}{(a − r)^2 + 2ar} \right |$
$ \lim_{r → a^-} \left | \frac{a^2-r^2}{a^2 + r^2} \right |$
$\left | \frac{a^2-a^2}{a^2 + r^2} \right | = 0$
Is this correct? This seems to simple too be right.
It is too easy (and incorrect). Your estimate seems to show that the maximum is $1$, independent of $\theta_\ast$. However, that is not true.
Let $s=r/a$. As $\theta_\ast\to0$, $$ \begin{align} \max_{r\lt a}\max_{|\theta|\ge\theta_\ast}\left|\frac{a^2-r^2}{(a−r)^2+2ar(1−\cos(\theta))}\right| &=\max_{r\lt a}\left|\frac{a^2-r^2}{(a−r)^2+2ar(1−\cos(\theta_\ast))}\right|\\ &=\max_{r\lt a}\left|\frac{a^2-r^2}{(a−r)^2+4ar\sin^2(\theta_\ast/2)}\right|\\ &=\max_{s\lt 1}\left|\frac{1-s^2}{(1−s)^2+4s\sin^2(\theta_\ast/2)}\right|\\ &\sim\frac1{2\sin(\theta_\ast/2)} \end{align} $$ when $s\sim1-2\sin(\theta_\ast/2)$.
For $2\sin(\theta_\ast/2)=\left(\color{#C00000}{\frac13},\color{#00A000}{\frac1{10}},\color{#0000F0}{\frac1{30}},\color{#C0A000}{\frac1{100}}\right)$