Find the limit of the following sequence or determine that the sequence diverges. \( a_{n}=-e^{-3 n} \cos n \)

To find the limit of the sequence \( a_{n} = -e^{-3 n} \cos n \), we analyze the behavior of the two components of the expression as \( n \) approaches infinity. First, consider the term \( e^{-3n} \). As \( n \to \infty \), this exponential term approaches 0 because the exponent \( -3n \) becomes very large and negative, which causes \( e^{-3n} \) to shrink towards 0. Next, the term \( \cos n \) oscillates between -1 and 1 for all values of \( n \). Hence, while \( \cos n \) does not converge to a specific value, it remains bounded. Combining these observations, the limit of the entire sequence can be determined as follows: \[ \lim_{n \to \infty} a_{n} = \lim_{n \to \infty} -e^{-3n} \cos n = -(\lim_{n \to \infty} e^{-3n})(\lim_{n \to \infty} \cos n). \] Since \( e^{-3n} \) tends towards 0, and \( \cos n \) is bounded, we apply the squeeze theorem: \[ -a_n = e^{-3n} |\cos n| \quad \text{where } 0 \leq |\cos n| \leq 1. \] Thus, \[ 0 \leq |a_n| = |-e^{-3n} \cos n| \leq e^{-3n}. \] As \( n \to \infty \), \( e^{-3n} \to 0 \), which implies \( |a_n| \to 0 \). This leads us to conclude: \[ \lim_{n \to \infty} a_{n} = 0. \] Therefore, the limit of the sequence \( a_n \) is \( \boxed{0} \).