If $f(x)=[x]\sin\frac{π}{[x+1]}$, then the points of discontinuity of f(x) in domain of the function are

∀ x ∈ I − {0}

f(x) is defined if [x + 1] ≠ 0 ⇒ 0 ≤ x + 1 < 1

i.e. [x + 1] = 0 if −1 ≤ x < 0. Thus dom f = R ~ [−1, 0).

We have $\sin\left(\frac{π}{[x+1]}\right)$ continuous at all points of R ~ [−1, 0) and [x] continuous on R ~ I, where I denote the set of integers. Thus the points where f can possibly be discontinuous are …….. , −3, −2, −1, 0, 1, 2, ……… But for 0 ≤ x < 1, [x] = 0 and $\sin\left(\frac{π}{[x+1]}\right)$ is defined.

Therefore , f(x) = 0 for 0 ≤ x < 1.

Also, f is not defined on [−1 , 0), so the continuity of f at 0 means continuity of f from right at 0. Since f is continuous from right at 0, f is continuous at 0. Hence the set of discontinuities of f is I ~ {0}.