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Let $Pin mathbb C$ 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.

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
$$
f_n(z) = hat f_n(z) = frac12 pi i int_gamma frachat f_n(zeta) , dzetazeta - z = frac12 pi i int_gamma fracf_n(zeta) , dzetazeta - z
$$
for $0 < |z - P | < rho$ and all $n$. And since $f_n to f$ uniformly on the image of $gamma$ it follows that
$$
f(z) = frac12 pi i int_gamma fracf(zeta) , dzetazeta - z
$$
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:
$$
a_k^(n) = frac12 pi iint_gamma fracf_n(z) , dz(z-P)^n+1
, , ,
a_k = frac12 pi iint_gamma fracf(z) , dz(z-P)^n+1
$$
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.