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

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,$g(n)=n-[n]$ = $n-n$ = o
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