Show that the function defined by

The given function is $g(x)=x-[x]$

It is evident that g is defined at all integral points.

Let $n$ be an integer.

Then,

$g(n)=n-[n]=n-n=0$

The left hand limit of f at x = n is,

$\lim _{x \rightarrow n^{-}} g(x)=\lim _{x \rightarrow n^{-}}(x-[x])=\lim _{x \rightarrow n^{-}}(x)-\lim _{x \rightarrow n^{-}}[x]=n-(n-1)=1$

The right hand limit of f at x = n is,

$\lim _{x \rightarrow n^{+}} g(x)=\lim _{x \rightarrow n^{+}}(x-[x])=\lim _{x \rightarrow n^{+}}(x)-\lim _{x \rightarrow n^{+}}[x]=n-n=0$

It is observed that the left and right hand limits of f at x = n do not coincide.

Therefore, f is not continuous at x = n

Hence, g is discontinuous at all integral points.