Question
Show that the function g(x) = x - [x] is discontinuous at all integral points. Here [x] denotes the greatest integer function.
✓
Answer
The given function is g(x) = x - [x]
It is evident that g is defined at all integral points.
Let n be an integer.
Then,
g(n) = n - [n] = n - n = 0
The left hand limit of f at x = n is,
$\lim\limits_{{\text{x}}\rightarrow\text{n}^-}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{n}^-}\big(\text{x}-[\text{x}]\big)=\lim\limits_{{\text{x}}\rightarrow\text{n}^-}(\text{x})-\lim\limits_{{\text{x}}\rightarrow\text{n}^-}[\text{x}]\\=\text{n}-(\text{n}-1)=1$
The right hand limit of f at x = n is,
$\lim\limits_{{\text{x}}\rightarrow\text{n}^+}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{n}^+}\big(\text{x}-[\text{x}]\big)\\=\lim\limits_{{\text{x}}\rightarrow\text{n}^+}(\text{x})-\lim\limits_{{\text{x}}\rightarrow\text{n}^+}[\text{x}]=\text{n}-\text{n}=0$
It is observed that the left and right hand limits of f at x = n do not coincide.
Therefore, f is not continuous at x = n
Hence, g is discontinuous at all integral points.
It is evident that g is defined at all integral points.
Let n be an integer.
Then,
g(n) = n - [n] = n - n = 0
The left hand limit of f at x = n is,
$\lim\limits_{{\text{x}}\rightarrow\text{n}^-}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{n}^-}\big(\text{x}-[\text{x}]\big)=\lim\limits_{{\text{x}}\rightarrow\text{n}^-}(\text{x})-\lim\limits_{{\text{x}}\rightarrow\text{n}^-}[\text{x}]\\=\text{n}-(\text{n}-1)=1$
The right hand limit of f at x = n is,
$\lim\limits_{{\text{x}}\rightarrow\text{n}^+}\text{g(x)}=\lim\limits_{{\text{x}}\rightarrow\text{n}^+}\big(\text{x}-[\text{x}]\big)\\=\lim\limits_{{\text{x}}\rightarrow\text{n}^+}(\text{x})-\lim\limits_{{\text{x}}\rightarrow\text{n}^+}[\text{x}]=\text{n}-\text{n}=0$
It is observed that the left and right hand limits of f at x = n do not coincide.
Therefore, f is not continuous at x = n
Hence, g is discontinuous at all integral points.
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