Show that the function defined by g (x) = x – [x] is discontinuou…

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 $x.$

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Answer

$g(x) = x - [x]$
Let $a$ be any integer.
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{-}}\text{g(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{-}}\left\{\text{x-[x]}\right\}$
$[$Put $ x - a - h, h > 0$ so that $h\rightarrow 0$ as $x\rightarrow a^-]$
$ = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{0}}\left\{\text{(a}-\text{h})-(\text{a}-\text{h})\right\} $
$= ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\left\{\text{(a}-\text{h})-(\text{a}-\text{1})\right\}$
$ = ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\text{(a}-\text{h}-\text{a}+1) $
$= \text{a}- 0-\text{a} + 1 = 1$
$^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{+}}\text{g(x)} = ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{+}}\left\{\text{x - [x]}\right\}$
$[$Put $x = a + h, h > 0$ so that $h\rightarrow 0$ as $x\rightarrow a^+]$
$= ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\left\{\text{(a}+\text{h})-(\text{a}+\text{h})\right\}= ^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\left\{\text{(a}+\text{h})-\text{a}\right\} =^{\ \ \text{Lt}}_{\text{h}\rightarrow\text{0}}\text{(h)}
= 0$
$\therefore\ ^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{-}}\text{f(x)}\neq^{\ \ \text{Lt}}_{\text{x}\rightarrow\text{a}^{+}}\text{f(x)}$
$\therefore f$ is discontinuous at $x = a$
But $a$ is any integral point.
$\therefore f $ is discontinuous at all integgral points.

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