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 $$x$$.

Solution

## The given function $$g(x)$$ is defined at all integral points.Let $$n$$ be an integer.Then $$g(n)=n-{n}=n-n=0$$$$L.H.L.$$ at $$x=n=\lim _{ x\rightarrow { n }^{ - } }{ g\left( x \right) } =\lim _{ x\rightarrow { n }^{ - } }{ \left( x-[x] \right) } =n-\left( n-1 \right) =1$$$$R.H.L.$$ at $$x=n=\lim _{ x\rightarrow { n }^{ + } }{ g\left( x \right) } =\lim _{ x\rightarrow { n }^{ + } }{ \left( x-[x] \right) } =n-n=0$$Since $$L.H.L.\neq R.H.L.$$Therefore $$g$$ is not continuous at $$x=n.$$ i.e., $$g(x)$$ is discontinuous at all integral points.MathematicsRS AgarwalStandard XII

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