Consider $$ X=\left\{\left.n+\frac{1}{n}\ \right\vert \ n\in\mathbb N^\ast\right\} $$ is the $\inf X= 2$? and $\sup X=+\infty$? Also, for $$M=\left\{\left.\frac{n+1}{n^2+1}\ \right\vert \ n\in\mathbb N\right\} \cup \left\{\frac{1}{\sqrt{2}}\right\}$$
is the $\inf M=0$? and $\sup M =1$?
If yes,how to prove the values.
For the first,
we have for $n>0$
$$n+\frac 1n=(\sqrt{n}-\frac{1}{\sqrt{n}})^2+2\ge 2$$
and $$2 = 1+\frac 11 \in X$$ thus $$\inf X =2$$
Given $A>0$, if we put
$$n = \lfloor A \rfloor +1$$ then $$ n +\frac 1n > n \ge A$$ thus $X$ is not bounded and $$\sup X =+\infty$$
For the second,
$$ 1=\frac{0+1}{0^2+1} \in X$$ and $$(\forall n\in \Bbb N) \;\; \frac{n+1}{n^2+1}\le 1 \implies \sup X=1$$
the sequence $(\frac{n+1}{n^2+1})$ is decreasing and converges to $0$, thus $$\inf X =0.$$