Let $\{\Sigma_n\}_{n \in \mathbb{N}}$ be a sequence of $C^\infty$, complete, properly embedded surfaces in $\mathbb{R}^3$ such that $0 \in \Sigma_n$ for every $n \in \mathbb N$ and $\{\Sigma_n\}_{n \in \mathbb{N}}$ is converging in the $C^k$ topology (with $k$ arbitrary large) on compact sets to a smooth complete surface $\Sigma$ passing through $0$.
I wonder what kind of property $\Sigma_\infty$ can inherit from the sequence. More precisely, if $\Sigma_n$ is the graphical, is it possible to show that $\Sigma_\infty$ is graphical as well?
Of course one needs to be careful. Indeed it is easy to construct a sequence of functions $f_n: \{(x, y, z) : z = 0\} \to \mathbb R$ such that $\Sigma_n := \text{graph}(u)$ satisfies the above properties but $\Sigma_\infty$ is not the graph of a function from $\{(x, y, z) : z = 0\} $. But one might still hope to prove that $\Sigma_\infty$ is graphical but w.r.t. another direction.
The example I have in mind is the following. Consider the function $f(x,y) = x^2 + y^2$. Consider the following sequence of points on its graph: $$ p_n \quad \colon = \quad (n, 0, n^2). $$ Consider now the sequence $$ \Sigma_n \quad \colon= \quad \text{graph}(f) - p_n. $$ Observe that every $\Sigma_n$ is a vertical graph and they are converging to a vertical plane, which is not a vertical graph, but still a graph w.r.t. to another direction.
Any help will be highly apreciated!
EDIT : I actually realized that it is possible to build a counter-example where $\Sigma_n$ are graphical, but $\Sigma_\infty$ is not. On the other hand I think to have a proof of the following:
weaker statement: If the $\Sigma_n$ are graphical, then the Gauss map of $\Sigma_\infty$ is contained in a closed hemisphere.