Math floor and limits? $\lim \limits_{n \to \infty}\frac{n}{2}\left\lfloor \frac{3}{n}\right\rfloor$

Question

I have these equations:

$$\lim \limits_{n \to \infty}\frac{n}{2}\left\lfloor \frac{3}{n}\right\rfloor$$

$$\lim \limits_{n \to \infty}\frac{2}{n}\left\lfloor \frac{n}{3}\right\rfloor$$

What is the difference between the two?

Answer 1

If $n\gt 3$ then $\left\lfloor \dfrac3n\right\rfloor=0$ because $\dfrac3n\lt1$. Thus the first is $0$.

The second one is equal to either $\dfrac23$, $\dfrac23\cdot\dfrac{k}{k+1}$, or $\dfrac23\cdot\dfrac{k}{k+2}$ where $n=3k+r$, with $k,r\in \mathbb Z$ integers and $r\lt3$. In the limit $n\rightarrow\infty$, $k\rightarrow\infty$, so the limit is $\dfrac23$.

Answer 2

For the first limit, $0<\frac3n<1$ for $n>3$, so $\left\lfloor \frac{3}{n}\right\rfloor=0$ for $n>3$. Therefore the first expression always equals zero and the limit is zero.

For the second limit, use the squeeze theorem (also called the sandwich theorem or pinch theorem) on

$$\frac{2}{n}\left( \frac{n}{3}-1\right) \le \frac{2}{n}\left\lfloor \frac{n}{3}\right\rfloor \le \frac{2}{n}\left( \frac{n}{3}\right)$$

to get a limit of $\frac23$.