About a sequence of holomorphic maps from annuli Let $A_n=\{z\in \mathbb{C}: \epsilon_n\leq |z| \leq 1\}$ and
$$
f_n\colon A_n\to \mathbb{C}, \quad n\in \mathbb{N},
$$ 
be a sequence of holomorphic functions such that $\lim_{n\to \infty} \epsilon_n=0$, and for any $r<1$, 
$\{f_n|_{\{z\in \mathbb{C}: r\leq |z| \leq 1\}}\}_{n\in \mathbb{N}}$  converges to $0$ in $C^\infty$-norm (Uniformly with all derivatives).
Is it true that either $\lim_{n\to \infty} ||f_n(x)||_{\infty} =0$ for all $x$ or (after passing to a subsequence) there are sequences of points $\{z_n\}_{n\in\mathbb{N}}$ and $\{w_n\}_{n\in \mathbb{N}}$ in the boundary components $\{|z|=\epsilon_n\}_{n\in \mathbb{N}}$ with 
$$
\Big(\lim_{n\to \infty} f_n(z_n) \neq  \lim_{n\to \infty} f_n(w_n) \Big) \in \mathbb{C}\cup \{+\infty\} \cup \{-\infty\}.
$$ 
 A: I am not sure that I understand your question correctly. In particular, I am not sure what $+\infty$ and $-\infty$ mean. 
As I first understood it, the answer is negative. Let $A(r,1)$ be the annulus of points with  $r<z<1$. 
Define $\newcommand{\eps}{\varepsilon}\eps_n := 1/n^2$, $A_n := A(\eps_n,1)$ 
and set 
$f_n\colon A_n \to \mathbb{C}; z\mapsto \frac{1}{nz}$.
Then clearly the functions converge uniformly to zero on every annulus $A(\eps,1)$. However, if $z_n$ is any sequence with $|z_n|=\eps_n$, then 
$$ |f_n(z_n)| = n \to \infty.$$
However, if by converging to $+\infty$ or $-\infty$, you mean that points should converge to infinity along different rays (e.g. asymptotically to the positive/negative real axis), I think the answer is positive.
Indeed, the maximum must be taken on the boundary, and because the function has no poles and takes values near zero, the image of the boundary cannot be contained in a small neighbourhood of some finite value. So, either you can find subsequences of values converging to finite values, or, for large $n$, the image of the boundary surrounds any fixed disc around the origin. In the latter case, it must contain arbitrarily large positive and negative real values. 
