show that the function f(x)=x-[x] is discountinious at all integral points

Question

this question is based on continuiity in this you have 2 chek tha function is discontuinious at all integral points .this qus is of 12 class

Answer 1

f(x)=x-[x] where [] means integer part. Let x be an integer:

For 0<h<1, f(x+h)=x+h-[x+h]=x+h-x=h; f(x-h)=x-h-[x-h]=x-h-(x-1)=x-h-x+1=1-h.

Therefore f(x+h)≠f(x-h). As h→0 the right limit and left limit are not the same therefore the limit of f(x) when x is an integer cannot be defined, making the function f discontinuous at integral points.

EXAMPLE

Let x=2 and h=0.1. f(2.1)=2.1-2=0.1; f(1.9)=1.9-1=0.9. 0.1≠0.9. So discontinuity.

Let x=2 and h=0.9. f(2.9)=2.9-2=0.9; f(1.1)=1.1-1=0.1. 0.9≠0.1. So discontinuity.