The degree of the minimal polynomial decreases to the limit $A \mapsto \mu_A$ (minimal polynomial of the matrix $A$) is not continuous.  
Take for example $A_n = \begin{pmatrix}
    0 & 1/n\\
    0 & 0\\
   \end{pmatrix}$.
But I think that:
if $(A_n)_{n \in \mathbb N}$ satisfies $A_n \to A$ (whatever the norm is since the dimension is finite) and if we suppose all $\mu_{A_n}$ and $\mu_A$ exist, then:
$\exists \ n_0, \ \forall \ n ≥ n_0,$ deg$(\mu_A) ≤ $deg$(\mu_{A_n})$
Do you have a hint to prove that?
 A: I bet this works: First show that (in any normed vector space) if $x_1,\dots,x_n$ are independent then there exists $\epsilon>0$ such that if $||x_j-y_j||<\epsilon$ for all $j$ then $y_1,\dots,y_n$ are independent. Apply that to $I,A,A^2,\dots$.
A: Assume that the required result is false.
There is a subsequence (noted again $(A_n)$) s.t. $(A_n)$ tends to $A$ and, for every $n$, $degree(\mu_{A_n})<degree(\mu_A)$. Considering a new subsequence, we may assume that $degree(\mu_{A_n})$ is a constant $1\leq r<degree(\mu_A)$, that is, $\mu_{A_n}(x)=x^r+\sum_{i=0}^{r-1} a_{n,i}x^i$.
On the other hand, the eigenvalues of $(A_n)$ are uniformly bounded (for $n$ large enough) by $\rho(A)+1$. Thus, the $(|a_{n,i}|)_{n,i}$ are bounded by a real $M>0$.
Considering a new subsequence, we may assume that, for every $i$, the sequence $(a_{n,i})_n$ converges to a real $a_i$. We deduce that the sequence $(\mu_{A_n})_n$ converges to the polynomial $\mu(x)=x^r+\sum_{i=0}^{r-1}a_ix^i$.
Finally, $\mu_{A_n}(A_n)=0$ converges to $\mu(A)=0$ and $\mu_A$ divides $\mu$, a contradiction.
