Is $F(x)=\frac{x}{x+1}$ for $x\in [0,1)$, $F(1)=\frac{1}{2}$ continuous?

Question

How is this function continuous? The book before me states that it's continuous.

$$ F(x)=\begin{cases} \dfrac{x}{x+1},&\;\;x\in [0,1),\\\\ \;\; \dfrac{1}{2},&\;\;x=1\end{cases} $$

I might be wrong

Let $\epsilon_0=\dfrac{1}{12},$ $x=\dfrac{1}{2}$ and $y=1$, then \begin{align}\left|F\left(\dfrac{1}{2}\right)-F\left(1\right)\right|&=\left|F\left(\dfrac{1}{2}\right)-F\left(1\right)\right|\\&=\left|F\left(\dfrac{1}{2}\right)-F\left(1\right)\right|\\&=\left|\dfrac{1}{3}-\dfrac{1}{2}\right|\\&=\dfrac{1}{6}>\dfrac{1}{12}=\epsilon_0\end{align} Then, it's not continuous. I know we can use Pasting lemma but $[0,1)\cap 1=\phi.$

Answer 1

Define$$\begin{array}{rccc}G\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&\dfrac x{x+1}.\end{array}$$Then $F=G$ and I suppose that it is clear to you that $G$ is continuous. Therefore, $F$ is continuous too.

On the other hand, there is no way of determining whther a function $\varphi\colon[0,1]\longrightarrow\mathbb R$ is continuous at $1$ just by using the values of $\varphi$ at $1$ and at $\frac12$, which is what you tried to do.

Answer 2

I like to think about it as a game between you and an adversary. You are tying to catch your adversary off guard by demanding that they produce a $\delta>0$ for any $\epsilon>0$ such that $0<|x-a|<\delta$ implies that $|f(x)-f(a)|<\epsilon$.

The definition of being continuous does not let you chose your $\delta$; that role is reserved for your adversary.

So when you chose $\epsilon_0=1/12$, you do not then also get to pick the $x$ and $y$ values. That is for the adversary to decide.