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.

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→0 as x→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→0 as x→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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