On singularity of limit holomorphic function Let $P\in \mathbb C$ and $f_n : D(P,r)\setminus \{P\} \to \mathbb C$ be a sequence of holomorphic functions such that there is a holomorphic function $f: D(P,r)\setminus \{P\} \to \mathbb C$ such that for every compact subset $A$ of $D(P,r)\setminus \{P\} $ , $\{f_n\}$ converges uniformly on $A$ to $f$. 
If $P$ is a removable singularity for every $f_n$, then is $P$ a removable singularity for $f$ also ? 
Let $\hat f_n$ be the extension of each $f_n$. By Cauchy Integral formula, I can see that $\lim_{n \to \infty} \hat f_n (P)$ exists. So a natural way to extend $f$ at $P$ would be to assign the value of this limit. But then I'm not sure whether the resulting function is holomorphic or not ... if the  $\hat f_n$ s would converge uniformly on compact sets in $D(P, r)$ to this extended $f$, we would be done, but I'm not sure whether that is true or not.
Please help. 
 A: Let $\gamma: [0, 2 \pi] \to \Bbb C$ be the circle $\gamma(t) = P + \rho e^{it}$ for some $\rho \in (0, r)$. Cauchy's integral formula holds for every extension $\hat f_n$ of $f_n$, therefore
$$
 f_n(z) = \hat f_n(z) = \frac{1}{2 \pi i} \int_\gamma \frac{\hat f_n(\zeta) \, d\zeta}{\zeta - z} = \frac{1}{2 \pi i} \int_\gamma \frac{f_n(\zeta) \, d\zeta}{\zeta - z}
$$
for $0 < |z - P | < \rho$ and all $n$. And since $f_n \to f$ uniformly on the image of $\gamma$ it follows that
$$
 f(z) = \frac{1}{2 \pi i} \int_\gamma \frac{f(\zeta) \, d\zeta}{\zeta - z}
$$
for $0 < |z - P | < \rho$. The right-hand side is a continuous (even holomorphic) function in $D(P,\rho)$, which means that $f$ can be holomorphically extended at $P$ as well.

Alternatively one can argue as follows: Each $f_n$ and $f$ can be developed into Laurent series at $z=P$:
$$
 f_n(z) = \sum_{k=-\infty}^\infty a_{k}^{(n)} (z-P)^k \, , \, 
f(z) = \sum_{k=-\infty}^\infty a_{k} (z-P)^k
$$
and the coefficients can be computed with a generalized Cauchy integral formula:
$$
a_{k}^{(n)} = \frac{1}{2 \pi i}\int_\gamma \frac{f_n(z) \, dz}{(z-P)^{n+1}}
\, , \, 
a_{k} = \frac{1}{2 \pi i}\int_\gamma \frac{f(z) \, dz}{(z-P)^{n+1}}
$$
 The locally uniform convergence implies that
$$
 \lim_{n \to \infty} a_{k}^{(n)} = a_k
$$
for all $k \in \Bbb Z$. If each $f_n$ has a removable singularity at $z=P$ then $a_{k}^{(n)} = 0$ for all $n$ and all $k < 0$, and consequently $a_k = 0$ for $k < 0$, so that $f$ has a removable singularity at $z=P$ as well.
