\( \lim _ { n \rightarrow \infty } \frac { ( - 1 ) ^ { n } } { 2 n + 5 } \)

To solve the limit \( \lim _ { n \rightarrow \infty } \frac { ( - 1 ) ^ { n } } { 2 n + 5 } \), we notice that as \( n \) approaches infinity, the denominator \( 2n + 5 \) approaches infinity as well. The numerator, \( (-1)^n \), oscillates between -1 and 1. As such, the absolute value of the fraction becomes \( \frac{1}{2n + 5} \), which tends to 0 as \( n \) increases. Due to the oscillating nature of the numerator, the limit can be evaluated as: \[ \lim_{n \to \infty} \frac{(-1)^n}{2n + 5} = 0 \] Thus, the limit is: \[ \boxed{0} \]