Let $\mathcal{A}\left(\mathbb{D}\right)$ denote the space of all holomorphic functions $f:\mathbb{D}\rightarrow\mathbb{C}$, where $\mathbb{D}$ is the open unit disk in $\mathbb{C}$, let $\omega$ be a positive real number, let $\left\{ f_{n}\right\} _{n\geq0}$ be a sequence in $\mathcal{A}\left(\mathbb{D}\right)$ converging uniformly on every compact subset of $\mathbb{D}$ to a limit $f\in\mathcal{A}\left(\mathbb{D}\right)$, and suppose that:$$\lim_{x\uparrow1}\left(1-x\right)^{\omega}f_{n}\left(x\right)=1,\textrm{ }\forall n\geq0$$
Is it then true that:$$\lim_{x\uparrow1}\left(1-x\right)^{\omega}f\left(x\right)=1$$ or, do I need more assumptions on the convergence of the $f_{n}$s (if so, what)? Also, note here that there is no guarantee that $f$ is holomorphic, or even finite-valued at $1$.
There is a sequence of polynomials $g_n$ that converge uniformly on compact subsets of $\mathbb D$ to the principal branch $g$ of $1/(1-x)^\omega$. Take $f_n = g + g_n$ and $f = 2 g$; you have $\lim_{x \uparrow 1} (1-x)^\omega f_n(x) = 1$ but $\lim_{x \uparrow 1} (1-x)^\omega f(x) = 2$.