I would need to show that the function $$f:\mathbb{R}\rightarrow \mathbb{R}~~~~~f(x)=0~\text{if} ~x<0, ~f(x)=2~\text{if} ~x\geq 0$$is not continuous at the point $x=0$ with the $\delta$-$\epsilon$ criteria .
Here's what I've been thinking about. Let $\epsilon =\frac{1}{2}$, let $n\in \mathbb{N}$ choose $\delta>0$. Then we know that there exists $n\in \Bbb{N}$ such that $\frac{1}{n}<\delta$, thus take $x=-\frac{1}{n}\in \mathbb{R}$ and notice that $|x|<\delta$. But at the same time it holds that $$|f(x)-2|=|-2|=2>\epsilon$$ Therefore $f$ is not continuous according to the $\delta$-$\epsilon$ criterion.
Remark alternativly one could take $x=0.5\delta$, this would also work.
Is it possible to do it this way?
Thank you very much for your answer.