Suppose that we have a sequence of analytic functions $f_n:D(0,\rho_n)\to\mathbb C$ where $(\rho_n)_n$ is a decreasing sequence of real numbers $>1$ that converges to $1$. Assume furthermore that $|f_n(z)|\leq M$ for all $z\in D(0, \rho_n)$ and all $n=1,2,\ldots$ Suppose that $f_n$ converges to an analytic function $f$ on the open unit disc, uniformly on compact subsets.
Question: Does $f_n$ converge (pointwise) on the unit circle?
No, consider sequence of functions $f_n(z) = (1-1/n)^{n}z^{n}$. Clearly, they are analytic on $D(0,1+1/n)$ and inside unit disk converge to $0$. Since $(1-1/n)^{n}$ has a limit, to show that they do not converge on the circle for any sub-sequence of integers one has to show that $e^{n_k\pi t}$ has no limit for almost all $t\in [0,2\pi]$. This is well known and easy to prove. Say, let's outline a proof that $\sin{n_k}t$ may have a limit on a set of measure zero only. Let $E$ be set of points where $\sin{n_k}t$ has limit and the limit is greater than $\epsilon > 0$. If $h$ is a characteristic function of this set then Fourier coefficient of $h$ tend to zero. At the same time from the definition of $E$ this limit is at least $|E|*\epsilon$ so $|E|$ must be zero. There is one more eventuality that the limit of $\sin{n_k}t$ is zero but this case is considered in the same way by looking at $\cos{n_k}t$ which (after passing to a sub-sequence if needed) is convergent to one (if $\sin{n_k}t$ tends to 0).