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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.
Solution
Let n ∈ I.
Then `lim_(x->n^-)[x] = n - 1`
∵[x] = n - 1 ∀ x ∈ [n - 1,n]
and g(n) = n - n = 0 ∵ [n] = n because n ∈ I]
Now,
`lim_(x->n^-) g(x) = lim_(x->n^-) (x - [x]) = lim_(x->n^-) x - lim_(x->n^-)[x] = n - (n - 1) = 1`
and `lim_(x->n^+) g(x) = lim_(x->n^+)(x - [x]) = lim_(x->n^+)x - lim_(x->n^+)[x] = n - n = 0`
Thus, `lim_(x->n^-) g(x) ne lim_(x->n^+)g(x)`
Hence, g is discontinuous at all integral points.
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