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

Here, g(x) = x - [x]

Let a be an integer, then [a-h]= a - 1, [a-h]=a and [a]=a

At x=a, LHL = $\underset{x\to a^{-}}{lim}g\left(x\right)=\underset{x\to a^{-}}{lim}\lambda (x-[x]$

Putting x=a-h as $x\to a^{-}$ when $h\to 0$

$\therefore \underset{h\to 0}{lim}(a-h-[a-h\left]\right)=\underset{h\to 0}{lim}[a-h-(a-1\left)\right]=\underset{h\to 0}{lim}[-h+1]=1(\therefore [a-h]=a-1)$

RHL = $\underset{x\to a^{+}}{lim}g\left(x\right)=\underset{x\to a^{+}}{lim}(x-[x\left]\right)$

Putting x=a-h as $x\to a^{+}$ when $h\to 0$

$\therefore \underset{h\to 0}{lim}(a+h-[a+h\left]\right)=\underset{h\to 0}{lim}(a+h-a)=\underset{h\to 0}{lim}h=0\therefore [a-h]=a$

$\therefore LHL\ne RHL$

Thus, g(x) is discontinuous at all intergral points. $\therefore [a-h]=a-1)$