We are given the matrix \( f(x) \): \[ f(x) = \begin{pmatrix} \cos x & 1 & 0 \\ 0 & 2 \cos x & 3 \\ 0 & 1 & 2 \cos x \end{pmatrix} \] We need to evaluate \( \lim_{x \to \pi} f(x) \). The limit of the matrix as \( x \to \pi \) can be found by taking the limit of each element in the matrix. Let's evaluate the limits:
The first element is \( \cos x \), and as \( x \to \pi \), \( \cos \pi = -1 \).
The second element is 1, which is constant.
The third element is 0, which is constant.
The fourth element is 0, which is constant.
The fifth element is \( 2 \cos x \), and as \( x \to \pi \), \( 2 \cos \pi = -2 \).
The sixth element is 3, which is constant.
The seventh element is 0, which is constant.
The eighth element is 1, which is constant.
The ninth element is \( 2 \cos x \), and as \( x \to \pi \), \( 2 \cos \pi = -2 \).
Thus, the limit of the matrix as \( x \to \pi \) is: \[ \lim_{x \to \pi} f(x) = \begin{pmatrix} -1 & 1 & 0 \\ 0 & -2 & 3 \\ 0 & 1 & -2 \end{pmatrix} \]
Therefore, the answer is \({-1} \), corresponding to option (A).