I have the following function:
$$F(x)=\begin{cases} \dfrac{x}{x+1},& \text{if }x\ge 0,\\ 0,& \text{otherwise}\end{cases}$$
I want to prove that it is right-continuous and non-decreasing.
To show it's right continuous, I had to show that $$\lim_{x\to 0^+} F(x)=F(0). $$ $$\lim_{x\to 0^+} F(x)= \frac{0}{1+0} = 0 \quad\mbox{ and }\quad F(0)=0.$$ Since they are both equal, that proves the right continuity. (correct me if I'm wrong plz)
For non-decreasing, I know that I must show that $x\le y$ implies $F(x)\le F(y)$, but I don't know how to do that. I can't find anything online to help.
2 Answers
$\begingroup$The usual way of proving that a function is non-decreasing is to analyze the sign of its first derivative: roughly, given a function $f$, it will be non-decreasing if $f'(x)\ge 0$.
Since your function is continuous and has no singularity, you just need to compute $F'$ and observe that it can never be negative. If you do not now how to do it, ask for some hints.
EDIT: The first derivative of $F$ is given by $$F'(x)=\begin{cases} \dfrac{1}{(x+1)^2},& x> 0,\\[5pt] 0,& x<0.\end{cases}$$
If you prefer not to use derivatives, you can also show that $F(x)\le F(y)$ if $x\le y$. First note that $\frac{x}{x+1}\ge 0$ for all $x\ge0$, so you just need to consider the case $0\le x\le y$. In that case, both $x+1$ and $y+1$ are positive, so: $$\frac{x}{x+1}\le\frac{y}{y+1} \Leftrightarrow x(y+1)\le y(x+1) \Leftrightarrow x \le y.$$
$\endgroup$ 3 $\begingroup$If you want to avoid using derivatives, which is the easiest way, try the following direct approach:
$$0<x<y: \frac x{1+x}<\frac y{1+y}\iff x+xy<y+xy\iff x<y$$
$\endgroup$
