$\det(\lim\limits_{n \to \infty} A_n)=\lim\limits_{n \to \infty}(\det A_n)$

Question

If $A$ is a $n \times n$ square matrix, $\det(\lim\limits_{n \to \infty} A_n)=\lim\limits_{n \to \infty}(\det A_n)$?

I think yes, since $\det(A_n)$ is a polynomial in the entries of the matrix, so it continuous. It suffices this argument to show the claim?