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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.

Mathematics
RS Agarwal
Standard XII

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