Let $f(x)=\begin{cases} 2x + 3&\text{ for }x\geq 1,\\ -x+5 &\text{ for }x<1. \end{cases}$
$f$ is continuous from the right at $x\geq1$. The proof would be:
Let $\epsilon>0$ be arbitrary.
Let $x_0\geq1$.
Let $\delta=\epsilon/2$.
Let $x\in R$ and $x_0\leq x<x_0+\delta$. Thus $x\geq1$.
Thus $|f(x)-f(x_0)|=|2x+3-2x_0-3|=|2x-2x_0|=2|x-x_0|<2\delta=2\epsilon/2=\epsilon$.
This comes from the definition for continuity from the right:$\forall\epsilon>0\; \exists\delta>0$ such that if $x\in I$ and $x_0\leq x<x_0+\delta$ then $|f(x)−f(x_0)|<\epsilon$.
Prove $f$ is discontinuous from the left at $x=1$ using the definition: $\exists\epsilon>0\; \forall\delta>0$ such that if $x\in I$ and $x_0-\delta<x\leq\ x_0$ then $|f(x)−f(x_0)|\geq\epsilon$.
I can't seem to find $\epsilon$. But I think the proof would go like:
Let $\epsilon$ = ?
Let $\delta >0$ be arbitrary.
Let $x_0<1$.
Let $x\in I$ that is to say $x<1$ and $x_0-\delta<x\leq\ x_0$
Then from there is figuring out $|f(x)−f(x_0)|\geq\epsilon$ which I don't get because $|f(x)−f(x_0)|=|-x+5 +x_0-5|=|-x+x_0|=|x-x_0|<\delta$.
So would you set $\epsilon\leq\delta$?
I know my definitions are correct. My teacher has drilled them into our brains. $I$ stands for the domain of $f$ which is the reals or $R$ except the domain is split in two. And by "from the right" and "from the left" I mean that the space between $x$ and $x_0$ denoted as $\delta$, or $|x-x_0|<\delta$, is only calculated on one side, either adding or subtracting $\delta$, not by doing both which would be $x_0-\delta<x<x_0+\delta$.