How to get the supremum and infimum of a set $$ \left\{ x \in\mathbb{R}\; \middle\vert\; \tfrac{x}{|x| + 1} < \tfrac{1}{3} \right\}$$
What is the supremum and infimum of this set? I thought the supremum is $\frac{1}{3}$. But can we say that for any set $ x < n$ that $n$ is the supremum of the set? And for the infimum I have no idea at all.  Also, let us consider this example:
$$  \left\{\tfrac{-1}{n} \;\middle\vert\; n \in \mathbb{N}_0\right\}$$
How can I find the infimum and supremum of this set? It confuses me a lot. I know that as $n$ gets bigger $\frac{-1}{n}$ asymptotically approaches $0$ and if $n$ gets smaller $\frac{-1}{n}$ approaches infinity, but that's about it. 

Considering separately cases $x\geqslant{0}$ and $x<0$ you can find range of $x$ for which $f(x)=\dfrac{x}{|x| + 1} - \dfrac{1}{3} <0.$
For $x\geqslant {0}$ $$f(x)=\dfrac{x}{x+1}-\dfrac{1}{3}=\dfrac{x+1-1}{x+1}-\dfrac{1}{3}=\dfrac{2}{3}- \dfrac{1}{x+1},$$
For $x<0$ $$f(x)=\dfrac{x}{1-x}-\dfrac{1}{3}=\dfrac{x-1+1}{1-x}-\dfrac{1}{3}= \dfrac{1}{1-x} -\dfrac{4}{3}  $$
therefore, the function $f(x)$ increases on $\mathbb{R}.$ Then  for $x \in [0, \,+\infty) $ the inequality $f(x)<0$ holds  for $0 \leqslant x < \dfrac{1}{2}.$
Solutions of the inequality $\dfrac{1}{1-x} -\dfrac{4}{3}<0$ in the second case lies in  $(- \infty, \, 0) \cap \left(- \infty, \, \dfrac{1}{4}\right)= (- \infty, \, 0) .$

Thus $f(x)<0$ for $x \in  (- \infty, \, 0)\cup \left[0, \, \dfrac{1}{2}\right)= \left(- \infty, \, \dfrac{1}{2} \right) \Rightarrow \sup \left \lbrace x\vert \; f(x)<0 \right \rbrace =  \dfrac{1}{2} ;  \;\; \inf \left \lbrace x\vert \; f(x)<0 \right \rbrace =  -\infty .$
