<s>[INST] Finding supremum & infimum of a given set 
If exist, find, $\sup S,\inf S$\begin{equation}
 S:=\left\{\left\lfloor\sqrt[|n|]{\frac{7}{3}}\right\rfloor\cdot
 \frac{n^{2}-2 n-4}{n^{2}-n-6}: n \in
 \mathbb{Z}\setminus\{-2,0,3\}\right\} \end{equation}

My attempt:
\begin{equation}f(n):=\left\lfloor\sqrt[|n|]{\frac{7}{3}}\right\rfloor\cdot \frac{n^{2}-2 n-4}{n^{2}-n-6}\end{equation}
$n=1, f(1)=\left\lfloor\frac{7}{3}\right\rfloor\cdot\frac{5}{6}=\frac{5}{3}$
$n=-1,f(-1)=\left\lfloor\frac{7}{3}\right\rfloor\cdot\frac{1}{4}=\frac{1}{2}$
$n=2,f(2)=\left\lfloor\sqrt{\frac{7}{3}}\right\rfloor\cdot\frac{1}{2}=1$
\begin{equation}\lim_{n\to -2}f(n)=\pm\infty=\lim_{{n\to 3}}f(n)\end{equation}
\begin{equation}\lim_{n\to\infty}f(n)=\left\lfloor\left(\frac{7}{3}\right)^{\left|\frac{1}{n}\right|}\right\rfloor\cdot\frac{1-\frac{2}{n}-\frac{4}{n^2}}{1-\frac{1}{n}-\frac{6}{n^2}}=1\end{equation}
$$\implies\inf S=\frac{1}{2},\;\sup S=\frac{5}{3}$$
Is this correct?
 [/INST] The points $n=0$, $n=-2$, and $n=3$ are excluded by definition. When $|n|\geq4$ the integer part of the square root is $1$ and the quotient is between $2/3 $ and $3/2$.
Besides the values you calculated, you have $f(-3)=11/6$. So the supremum is actually a maximum, and it is equals $11/6$. The infimum is a minimum and it equals $1/2$.
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