Find the limit of the following sequence or determine that the sequence diverges. \( \left\{b_{n}\right\} \) if \( b_{n}=\left\{\begin{array}{ll}\frac{n}{n+6} & \text { for } n \leq 4600 \\ n e^{-n} & \text { for } n>4600\end{array}\right. \) Select the correct choice below and fill in any answer boxes to complete the choice. A. The limit of the sequence is \( \square \). (Type an exact answer.) B. The sequence diverges.

To determine the limit of the sequence \( b_n \), we will analyze the two cases given: 1. For \( n \leq 4600 \), the sequence is defined as \( b_n = \frac{n}{n+6} \). As \( n \) approaches \( 4600 \), this expression approaches \( \frac{4600}{4606} \), which simplifies to \( \frac{4600}{4606} = \frac{2300}{2303} \), approximately equal to \( 0.999 \). 2. For \( n > 4600 \), we have \( b_n = n e^{-n} \). As \( n \) grows larger, the term \( e^{-n} \) approaches \( 0 \) much faster than \( n \) approaches infinity. Hence, \( n e^{-n} \) approaches \( 0 \). Now, we observe the limit for large \( n \). For \( n \to \infty \), \( b_n \to 0 \). Thus, the behavior of the sequence depends on which case we are analyzing. As \( n \to 4600 \), the first case results in \( \frac{2300}{2303} \); however, for \( n > 4600 \), the sequence clearly trends down to \( 0 \). Since the sequence converges to \( 0 \) as \( n \) approaches infinity, we can conclude: A. The limit of the sequence is \( 0 \).