A special limit for producting of eigenvalues

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Let (k>0) $A_{k},A$ be complex matrices with size $n\times n$ such that $\lim_{k}A_{k}=A$. Let $f$ be a function that $f(A)=r_{1}^{m_{1}}...r_{t}^{m_{t}}$ where $r_i$'s are eigenvalues of $A$ and $ m_{i} $'s are algebraic multiplicities of $r_i$'s.

Then we want to prove $\lim_{k}f(A_{k})=f(A )$. How?

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Product of eigenvalues of a square matrix equals determinant of that matrix, so that $f$ is continuous. Apply some theorem you know about limit of continuous functions.