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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 Mark
Question
Show that the function defined by $g(x) = x - \left[ x \right]$ is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.
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Answer

Let n $\in$ $I$
Then,$\mathop {\lim }\limits_{x \to {n^- }}[x]=n-1$
$\therefore [x]=n-1\ \forall \ x\in [n-1,n] $
and g(n) = n - n = 0 $[\therefore [n]=n\ because\ n\in I]$
Now, $\mathop {\lim }\limits_{x \to {n^-}} g(x)=\mathop {\lim }\limits_{x \to {n^-}}(x-[x])=\mathop {\lim }\limits_{x \to {n^- }}x-\mathop {\lim }\limits_{x \to {n^- }}[x]$
$=n-(n-1)=1$
Also, $\mathop {\lim }\limits_{x \to {n^+ }}g(x)=\mathop {\lim }\limits_{x \to {n^ + }}(x-[x])$
$=\mathop {\lim }\limits_{x \to {n^ + }}x-\mathop {\lim }\limits_{x \to {n^ + }}[x]=n-n=0$
Thus, $\mathop {\lim }\limits_{x \to {n^ - }}g(x)\ne \mathop {\lim }\limits_{x \to {n^ + }} g(x)$
Hence, g(x) is discontinuous at all integral points.

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