Answer
$f$ is discontinuous at $x=1$.
Work Step by Step
$x=1$ and $x=3$ are of interest to us, as they are the only values where $f(x)$ can have a discontinuity.
At $x=1$:
Left-sided limit:$\qquad \displaystyle \lim_{x\rightarrow 1^{-}}f(x)=1-1=0$
Right-sided limit:$\qquad \displaystyle \lim_{x\rightarrow 1^{+}}f(x)=1+2=3$
The limit does not exist, so $f$ is discontinuous at 1.
At $x=3$:
Left-sided limit:$\qquad \displaystyle \lim_{x\rightarrow 3^{-}}f(x)=3+2=5$
Right-sided limit:$\qquad \displaystyle \lim_{x\rightarrow 3^{+}}f(x)=3^{2}-4=5$
The limit exists, $L=5.$
Function value:$\quad f(3)=3^{2}-4=5$
Thus $f$ is continuous at $3$.
