Case 2 Let c be an integer Then we can find a sufficiently small real number
r > 0 such that [c – r] = c – 1 whereas [c + r] = c This, in terms of limits mean that
xlim
→c−
f (x) = c – 1, lim
x
→c+
f (x) = c
Since these limits cannot be equal to each other for any c, the function is
discontinuous at every integral point 5