Show that the function g (x) = x − [x] is discontinuous at all integral points

Given: $\mathrm{g}(x)=x-[x]$

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

Let $n \in Z$.

Then,

$\mathrm{g}(n)=n-[n]=n-n=0$

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,

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

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

i.e. $\lim _{x \rightarrow n^{-}} g(x) \neq \lim _{x \rightarrow n^{+}} g(x)$

So, $f$ is not continuous at $x=n, n \in Z$

Hence, $g$ is discontinuous at all integral points.