A real function f is said to be continuous at x = c, where c is any point in the domain of f if :
where h is a very small ‘+ve’ no.
i.e. left hand limit as x → c (LHL) = right hand limit as x → c (RHL) = value of function at x = c.
This is very precise,
We can summarise it as,
A function is continuous at x = c if :
Given,
g(x) = x – [x] ……equation 1
We need to prove that g(x) is discontinuous at every integral point.
Note : Idea of greatest integer function [x] –
Greatest integer function can be seen as an input output machine in which if you enter a number, It returns the greatest integer that is just less than number x.
For example :
[2.5] = 2 ; [9.99998] = 9 ; [–3.899] = –4 ; [4] = 4
To prove g(x) discontinuous at integral points we need to show that
Where c is any integer.
Let c is any integer
∴ g(c) = c – [c]
= c – c = 0 [using eqn 1]
[∵ c is integer and h is very small positive no, so (c–h) is a number less than integer c
∴ greatest integer less than (c –h) = (c–1)]
[using eqn 1 and idea of gif function]
Thus,
LHL ≠ RHL
∴ g(x) is discontinuous at every integral point.
