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. 
 A: For $x\geq 0$, the condition $\frac{x}{|x|+1}<1/3$ is equivalent to $x<x/3+1/3$, which in turn is equivalent to $x<1/2$.
So the supremum of your set is $1/2$.
For $x\leq 0$, we have that $\frac{x}{|x|+1}<1/3$ is equivalent to $x<-x/3+1/3$, and then to $4x<1$, which is true for every nonpositive number.
So the infimum of your set is $-\infty$.
Another way to put it is to observe that the set under consideration is $]-\infty,1/2[$.
For the second set $\{-1/n|n\geq 1\}=\{-1,-1/2,-1/3,\ldots\}$, the infimum is a mimimum and is equal to $-1$, while the supremum is the limit of this increasing sequence, namely $0$.
A: Let $x$ be negative. then $\dfrac{x}{|x|+1}$ is negative, and in particular $\lt \dfrac{1}{3}$.
So there is no infimum. (But some people allow the symbol $-\infty$ as an infimum.)
For the supremum, note that there are positive $x$ such that $\dfrac{x}{x+1}\lt \dfrac{1}{3}$.
Since $\dfrac{x}{x+1}=1-\dfrac{1}{x+1}$, our function is increasing for positive $x$. Solve $1-\dfrac{1}{x+1}=\dfrac{1}{3}$. We get $x=\dfrac{1}{2}$. 
Thus $\dfrac{1}{2}$ is an upper bound for our set.
However close we are to $\dfrac{1}{2}$, but below $\dfrac{1}{2}$, we will have $\dfrac{x}{x+1}\lt\dfrac{1}{3}$. So there is no cheaper upper bound than $\dfrac{1}{2}$. 
Remark: For your other question, I think you intend to look at $\{-\frac{1}{n}\}$, where $n$ ranges over the positive integers. The smallest element of this set is $-1$. It is therefore the infimum. For the supremum, note that our numbers are all $\lt 0$, but can be made arbitrarily close to $0$ by choosing $n$ large enough. So the supremum is $0$.
A: 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 $x \in  (- \infty, \, 0)\cup \left[0, \, \dfrac{1}{2}\right)= \left(- \infty, \, \dfrac{1}{2} \right) \Rightarrow \sup \left \lbrace x\vert \; f(x)<0 \right \rbrace =  \dfrac{1}{2} ;  \;\; \inf \left \lbrace x\vert \; f(x)<0 \right \rbrace =  -\infty .$
A: Hint: Find the range of 
$$f(x)=\frac{x}{\left|x\right|+1}-\frac13$$
This function is continuous and increasing.
