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