Show that the function defined by


Show that the function defined by $g(x)=x-[x]$ is discontinuous at all integral point. Here $[x]$ denotes the greatest integer less than or equal to $X$.


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

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

Let $n$ be an integer.



The left hand limit of 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 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.

Thereforef is not continuous at x = n

Hence, g is discontinuous at all integral points.



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