Showing $f_j(0)$ converges to $f(0)$ where $f_j$ and $f$ are rational functions. Let $A=\{z \in \mathbb{C}: \frac{1}{2} < |z| < 1 \}$. Let $\{f_j\}_{j=1}^\infty$ be a sequence of rational functions and let $f$ be a rational function. Suppose none of these functions have poles on $A \cup \{0 \}$, $f_j$ converges uniformly to $f$ on $A$, and each $f_j$ is nonzero on the closed unit disc. I am trying to show that $f_j(0) \rightarrow f(0)$.
I am having trouble with this because it is possible that even though each $f_j$ is nonzero, the numerator has a zero not in the closed unit disc. Otherwise, I would be able to assume the numerator was a constant and perhaps proceed from there. I would appreciate any help with how to start this proof.
 A: We need the extra condition that $f$ is not identically zero since otherwise the result doesn't hold as $f_n(z)=\frac{1}{(4z)^n+1}$ satisfies $f_n(0)=1, f_n(z) \to 0$ uniformly on A, $f_n$ rational with no poles on $A$ or at $0$ etc
Local uniform convergence on $A$ and the rest of the hypothesis is enough to conclude that $f_j \to f$ locally uniformly on the open disc; in particular, since $f_j, f$ have no poles at zero, it follows $f_j(0) \to f(0)$
(where local uniform convergence is meant in the sense of meromorphic functions, so if $w$ is not a pole of $f$ then there is a neighborhood $W$ and an $n(W)$ for which $f_n$ has no pole in $W$ for $n \ge n(W)$ and $f_n \to f$ uniformly on $W$ while if $w$ is a pole of $f$, poles of $f_n$ accumulate there with the corresponding order and $1/f_n \to 1/f$ in a neighborhood of $w$ - so for example if $w$ is a pole of order $2$, then for every small enough neighborhood, there is $n(W)$ for which $f_n$ has no zeroes and either two simple poles or a double pole there for $n \ge n(W)$ and $1/f_n \to 1/f$ uniformly in $W$)
Proof: $f$ not identically zero as noted so by Hurwitz $f$ has no zeroes in $A$ and $1/f_n \to 1/f$ locally uniformly in $A$; but if $g_n =1/f_n, g=1/f$, the hypothesis on the zeroes of $f_n$ implies that $g_n$ is holomorphic in the open unit disc, $g_n \to g$ locally uniformly in $A$;
in particular $g_n$ is uniformly locally bounded in $A$ so by maximum modulus ($\sup_{|z| \le 1-\epsilon}|g_n(z)|=\sup_{|z| = 1-\epsilon}|g_n(z)|$) it follows that $g_n$ is locally uniformly bounded in the opne unit disc so it is a normal family;
but now every subsequence $s$ of $g_n$ on the open unit disc has a subsequence converging to a holomorphic function $h_s$ on the open unit disc that is $g$ on $A$, hence by the identity principle, $h_s=h$ for all subsequences of $g_n$, $g_n \to h$ and $h=g$ on $A$, while $h$ is holomorphic on the open unit disc; since $g$ is meromorphic on the open unit disc (rational), it follows that $g=h$ and $1/f$ is holomorphic in the open unit disc too, while $1/f_n \to 1/f$ locally uniformly there and we are done (since that clearly implies $f_n \to f$ in the sense described above)
