Solutions

(a) Intuitively, $a_{n} = \lfloor \frac{5}{n} \rfloor$ seems to converge to 0. Notice that we can define $a_{n}$ piece-wise, to obtain an insight. That is,

$$a_{n} = \lfloor \frac{5}{n} \rfloor = \begin{cases} 5 & n = 1 \\ 2 & n = 2 \\ 1 & 3 \leq n \leq 5 \\ 0 & n \geq 6 \end{cases}$$

Thus, for every $\epsilon > 0$, choose $N = 6$. We then have

$$\forall n \in \mathbb{N}, \quad n \geq N \implies 0 = \lfloor \frac{5}{n} \rfloor < \epsilon.$$

(b) Intuitively, $a_{n} = \lfloor \frac{12 + 4n}{3n} \rfloor$ converges to 1 since

$$b_{n} = \frac{12 + 4n}{3n}$$

converges to $\frac{4}{3}$. In fact, we realize that

$$\forall n \geq 7, \quad 1 < b_{n} < 2.$$

Thus, for every $\epsilon > 0$, choose $N = 7$. We then have

$$\forall n \in \mathbb{N}, \quad n \geq N \implies 0 = \abs{\lfloor \frac{12+4n}{3n}\rfloor - 1} < \epsilon.$$