Answer
The function has a discontinuity at each integer value of $x.$
Work Step by Step
To prove let $n$ be any integer and let $z=n-8$:
$\lim\limits_{x\to n^-}f(x)=[[n^--8]]=z^-\to (z-1)\lt z^-\leq z\to$
$\lim\limits_{x\to n^-}f(x)=z-1.$
$\lim\limits_{x\to n^+}f(x)=[[n^+-8]]=z^+\to z\lt z^+\leq (z+1)\to$
$\lim\limits_{x\to n^+}f(x)=z.$
Since $\lim\limits_{x\to n^-}f(x)\ne\lim\limits_{x\to n^+}f(x)\to\lim\limits_{x\to n}f(x)$ does not exist and hence the function is not continuous at $n.$ These are non-removable discontinuities.
