I am not sure how to approach these supremum and infimum sets. I also have to give a min or max
$$A = \Big\{(-1)^n+\frac{1}{n+1}\ :\ n\in \Bbb{N}\Big\}$$ $$B = \Big\{\frac{x}{x+1}\ :\ x\in [0,\infty)\Big\}$$ $$C = \bigcup_{n=1}^\infty\Big[\frac{1}{2n},\frac{1}{2n-1}\Big]$$
Any help is much appreciated!
For $A$, note that $\frac{1}{n+1}\xrightarrow{n\to\infty} 0$ and furthermore, it is monotonically decreasing. $(-1)^n$ just cycles between $-1$ and $1$.
For $n = 1$ we obtain $(-1)^1 + \frac12 = -\frac12$.
For $n=2$ we obtain $(-1)^2 + \frac13 = \frac43$.
Now, notice that for $n \ge 2$ we have:
$$(-1)^n + \frac1{n+1} \le 1 + \frac13 = \frac43$$
Since the value for $n = 1$ is smaller than $\frac43$ we conclude that $\frac43$ is the supremum of $A$. Since it is attained for $n = 2$, it is also the maximum of $A$.
To obtain the infimum, notice that $A$ is bounded from below by $-1$:
$$-1 < (-1)^n + \frac{1}{n+1}$$
Futhermore, elements of $A$ approach arbitrarily close to $-1$. To show this, take $\varepsilon > 0$. Since $\frac{1}{n+1}\xrightarrow{n\to\infty} 0$, there exists an odd integer $n_0 \in \mathbb{N}$ large enough so that $\frac1{1+n_0} < \varepsilon$.
We have:
$$(-1)^{n_0} + \frac1{n_0 + 1} = -1 + \varepsilon$$
We conclude that $\inf A = -1$. It is not a minimum because $\frac1{n+1} > 0$ so it cannot be attained for any $n \in \mathbb{N}$.
For $B$, we have $\frac{x}{x+1} \ge 0$ and for $x = 0$ obviously $\frac{0}{0+1} = 0$. Thus, $\min B = 0$.
$B$ is bounded from above by $1$:
$$\frac{x}{x+1} \le \frac{x+1}{x+1} = 1$$
Now recall that $\lim_{x\to\infty} \frac{x}{x+1} = 1$, so there exists $x > 0$ large enough so that $\frac{x}{x+1} > 1- \varepsilon$ for any $\varepsilon > 0$. Thus, $\sup B = 1$. It is not a maximum since:
$$\frac{x}{x+1} < \frac{x+1}{x+1} = 1$$
Can you manage $C$ on your own?