Question
Find all the points of discontinuity of the greatest integer function defined by f(x) = [x], where [x] denotes the greatest integer less than or equal to x.
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Answer
First, observe that f is defined for all real numbers. The graph of the function is given in the figure. From the graph, it looks like that f is discontinuous at every integral point. Below we explore if this is true.
Case 1: Let c be a real number which is not equal to any integer. It is evident from the graph that for all real numbers close to c the value of the function is equal to [c]; i.e., $\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} \,[x] = [c]$. Also f(c) = [c] and hence the function is continuous at all real numbers not equal to integers.
Case 2: Let c be an integer. Then we can find a sufficiently small real number r > 0 such that [c – r] = c – 1 whereas [c + r] = c.
This, in terms of limits mean that
$\mathop {\lim }\limits_{x \to {c^ - }} f(x) = c - 1,\mathop {\lim }\limits_{x \to {c^ + }} f(x) = c$
Since these limits cannot be equal to each other for any c, the function is discontinuous at every integral point.
Case 1: Let c be a real number which is not equal to any integer. It is evident from the graph that for all real numbers close to c the value of the function is equal to [c]; i.e., $\mathop {\lim }\limits_{x \to c} f(x) = \mathop {\lim }\limits_{x \to c} \,[x] = [c]$. Also f(c) = [c] and hence the function is continuous at all real numbers not equal to integers.
Case 2: Let c be an integer. Then we can find a sufficiently small real number r > 0 such that [c – r] = c – 1 whereas [c + r] = c.
This, in terms of limits mean that
$\mathop {\lim }\limits_{x \to {c^ - }} f(x) = c - 1,\mathop {\lim }\limits_{x \to {c^ + }} f(x) = c$
Since these limits cannot be equal to each other for any c, the function is discontinuous at every integral point.
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