Vector valued analytic functions near essential singularity Suppose $f:\mathbb{C}\setminus\{0\}\to X$ is a vector valued analytic function, that has an essential singularity at $0$ ($X$ is some Banach space). It can be easily shown that, in this case, $f$ must be unbounded near $0$. I am intrested whether a stronger condition holds, namely, is it true that for any sequence $z_n\to 0$, we have that $||f(z_n)||\to\infty$? 
This is not true in the $\mathbb{C}$-valued setting. Indeed, by Picard's theorem, such a  $f:\mathbb{C}\setminus\{0\}\to \mathbb{C}$, while still unbounded near $0$, will attain any value, except perhaps one, in any neighborhood of $0$. Thus this function will be constant on many sequences $z_n$ converging to $0$. However, there is no Picard in the vector-valued setting, so perhaps there is a chance that the stronger requirement will hold?  
Edit: The situation I am interested is more concrete, when $f(z)=(zI-T)^{-1}x$, where $T$ is quasinilpotent and injective, and $x$ is some fixed non-zero vector. 

This need not be the case.
We could have a function of the form $f(z) = g(z)\cdot x$ where $g \colon \mathbb{C}\setminus \{0\} \to \mathbb{C}$ is holomorphic with an essential singularity at $0$, and $x \in X \setminus \{0\}$.
But it can be the case, e.g. if $f(z) = g(z)\cdot x + h(z)\cdot y$, where $x$ and $y$ are linearly independent, $g$ has an essential singularity at $0$ and $h$ has a pole at $0$.
