Basic Idea:

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, using our fundamental idea of limit from class 11 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]

LHL =

[∵ 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)]

RHL = [using eqn 1 and idea of gif function]

Thus, LHL ≠ RHL

∴ g(x) is discontinuous at every integral point.