Solution: We need to find $K \in \mathbb{N} $ such that for any $k \geq K$ \begin{align*} 3 - \varepsilon \lt \frac{3k + 1}{k + 3} \lt 3 + \varepsilon. \end{align*} At first we will find $K$ such that for any $k \geq K$ \[ 3 - \varepsilon \lt \frac{3k + 1}{k + 3} \] is true. So, \begin{align*} 3 - \varepsilon \lt \frac{3k + 1}{k + 3} & \iff (3 - \varepsilon ) (k + 3) \lt 3k + 1 \\ & \iff 9 - k \varepsilon - 3\varepsilon \lt 1 \\ & \iff 8 \lt \varepsilon (k + 3) \\ & \iff k \gt \frac{8}{\varepsilon } - 3. \end{align*}

From the Archimedean property, we choose $K \in \mathbb{N} $ such that $K > \frac{8}{\varepsilon } - 3$. So for any $k \geq K_1$, we have \[ k > \frac{8}{\varepsilon } - 3 \implies 3 - \varepsilon \lt \frac{3k + 1}{k + 3}. \] On the other hand, note that for any $k \in \mathbb{N} $ \[ 3 - \frac{3k + 1}{k + 3} = \frac{3k + 3 - 3k - 1}{k + 3} = \frac{2}{k + 3} > 0. \] Thus, for any $k \in \mathbb{N} $ we have \[ \frac{3k + 1}{k + 3} \lt 3 + \varepsilon . \] Thus, for any $k \geq K$ we have \[ 3 - \varepsilon \lt \frac{3k + 1}{k + 3} \lt 3 + \varepsilon. \]