Continuity and Differentiability — Class 12 Maths Solution

Let $n \in I.$
Then,$\mathop {\lim }\limits_{x \to {n^ - }} [x] = n - 1$ [
and $g(n) = n - n = 0.$ [ [n] $=$ n because $n \in I$]

Now, $\mathop {\lim }\limits_{x \to {n^ - }} g(x) = \mathop {\lim }\limits_{x \to {n^ - }} (x - [x]) = \mathop {\lim }\limits_{x \to {n^ - }} x - \mathop {\lim }\limits_{x \to {n^ - }} [x]$
$= n - (n - 1) = 1$

and $\mathop {\lim }\limits_{x \to {n^ + }} g(x) = \mathop {\lim }\limits_{x \to {n^ + }} (x - [x])$
$= \mathop {\lim }\limits_{x \to {n^ + }} x - \mathop {\lim }\limits_{x \to {n^ + }} [x] = n - n = 0$

Thus, $\mathop {\lim }\limits_{x \to {n^ - }} g(x) \ne \mathop {\lim }\limits_{x \to {n^ + }} g(x).$

Hence we can say that g(x) is discontinuous at all integral points.