Fix $m\in \mathbb{N}.$
For each $n\in \mathbb{N},$ let $A\_n\in \mathbb{M}\_{m}(\mathbb{C}),$ $X\_n\in \mathbb{C}^m,$ and $B\_n\in \mathbb{C}^m.$ Suppose that
$$X\_{n+1}=A\_n X\_n+B\_n,$$
$$\lim\_{n\rightarrow \infty} A\_n=A\_0,$$
$$|\det(A\_0)|>1,$$
the moduli of all entries of $A\_0$ are greater than 1,
and
$$\limsup\_{n\rightarrow \infty}\|B\_n\|^{1/n}<1,$$
where $\|\cdot\|$ is the Euclidean norm.


Is it possible that
$$\limsup\_{n\rightarrow \infty}\|X\_n\|^{1/n}=1?$$
I have the feeling that the answer should be "No." But I do not know how to prove it.
For $m=1,$ it is not difficult to show that the answer is no. Could somebody suggest how to prove or disprove it? Any references (books or papers) are very welcome? Thank you.



Vector recurrences (asymptotic property)
$\newcommand\1{\mathbf1}\newcommand\R{\mathbb R}$Yes, this is of course possible. E.g., this will be so if for all $n$ the matrix $A\_n$ is the diagonal matrix with $1,2\dots,2$ on the diagonal, $B\_n=0$, and $X\_n=[1,0,\dots,0]^\top$.




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The OP has changed the question, thereby invalidating the above answer. After the change, the answer is still the same: yes, this is possible. E.g., let $m=2$ and for all $n$ let $A\_n:=\begin{bmatrix}3&2\\ 2&3\end{bmatrix}=:A\_0$ and $B\_n:=0$.


Then all your conditions on $A\_n$ and $B\_n$ hold. However, $1$ is an eigenvalue of $A\_n=A\_0$ for all $n$.
It remains to let $X\_1$ be an eigenvector corresponding to the eigenvalue $1$ of $A\_n$; say, we can let $X\_1:=[-1,1]^\top$.


More generally, for any $m\ge2$ and for all $n$ we can let $A\_n=A\_0$ have (say) $3$ everywhere on the diagonal and $2$ everywhere off the diagonal, so that the eigenvalues of $A\_n=A\_0$ are $1$ (of multiplicity $m-1$) and $1+2m$ (so that $\det A\_0=1+2m>1$). Indeed, the vector $\1:=[1,\dots,1]^\top\in\R^m$ is an eigenvector of $A\_n=A\_0$ corresponding to the eigenvalue $1+2m$, and all nonzero vectors in $\R^m$ orthogonal to $\1$ are eigenvectors of $A\_n=A\_0$ corresponding to the eigenvalue $1$.


