Given a function $f\colon [0,1] \cup (2, 3] \to [0, 2]$ defined by 

$$f(x)=\begin{cases} x & 0 \le x \le 1 \\ x-1 & 2 < x \le 3 \end{cases}$$

My question is that is $f$ is continuous or not continuous ?

My attempt : Suppose if i take $c=0$ , then for every $\epsilon >0$ , there is exist $\delta$ such that $|x-c| < \delta$ such that $|f(x)-f(c)| <\epsilon$

But here $\lim\_{ x\rightarrow 0^+}f(x)= -1$ and $\lim\_{x\rightarrow 0^{-}} = 0$

So $f$ is not continuous