Calculus 10th Edition

The function has a discontinuity at each integer value of $x.$
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.