Given function is (x) = [x], where [x] denotes the greatest integer less than or equal to 

. i.e., f(x) = [x] = f(x) = \(f(n) = \begin{cases} x & \quad x \,is\, an\, integer\\ x-1 & x \,is\, not\, an\, integer \end{cases}\)

Since, both functions x and x − 1 are continuous function. (Because, all polynomial functions are continuous and both functions x and x − 1 are polynomial of degree 1. ) 

Then, only doubtful points are integers. 

Let us discuss about the continuity of function f(x) at arbitrary integer x = c.

The left hand limit of function f(x) at x = c is f(c −) = \(\lim\limits_{x \to c^-} f(x)\)  = \(\lim\limits_{h \to 0} f(c - h)\) 

= \(\lim\limits_{h \to 0} [c - h]\) = c.

 (Because, c is an integer, therefore, c + h > c and not an integer, therefore, [c + h] = c )

Therefore, f(\(c^-\)) ≠ f(\(c^+\)) means the left hand limit and the right hand limit of function f(x) = [x] at x = c is not equals. 

Therefore, the function f(x) = [x] is not continuous at x = c and c is an arbitrary integer. 

Therefore, the function (x) = [x] is not continuous at any integer.

Hence, the function f(x) = [x] is discontinuous at all integers.