Question

Show that the function defined by g(x) = x − [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.

Answer 1

Let n ∈ I.

Then `lim_(x->n^-)` [x] = n − 1  ....(∵ [x] = n − 1 ∀ x ∈ [n − 1, n])

and g(n) = n − n = 0  ...(∵ [n] = n because n ∈ I])

Now,

`lim_(x->n^-) g(x) = lim_(x->n^-) (x - [x]) = lim_(x->n^-) x - lim_(x->n^-)[x]`

= n − (n − 1) = 1

and `lim_(x->n^+) g(x) = lim_(x->n^+)(x - [x]) = lim_(x->n^+)x - lim_(x->n^+)[x]`

= n − n = 0

Thus, `lim_(x->n^-) g(x) ne lim_(x->n^+)g(x)`

Hence, g is discontinuous at all integral points.