If f:R→R is a function defined by f(x) = [x]cos(2πx), where [x] denotes the greatest integer function, then f is:

For n∈I, limx→n+f(x) = limx→n+[x]cos(2πx) = ncos(2πn) = 0
limx→n-f(x) = limx→n-[x]cos(2πx) = (n-1)cos(2πn) = n - 1
Since limx→n+f(x) ≠ limx→n-f(x), f(x) is discontinuous at all integers except at x = 0.
At x = 0, limx→0+f(x) = limx→0+[x]cos(2πx) = 0
limx→0-f(x) = limx→0-[x]cos(2πx) = -1cos(0) = -1
Since limx→0+f(x) ≠ limx→0-f(x), f(x) is discontinuous at x = 0.
Therefore, f(x) is discontinuous only at non-zero integral values of x.