We have :f(x)= .................................................. .........................................1

AND now we have the following cases:

1) |x|<1 <===> -1<x<1 ,then .

2) x=1 ,then .

3) x>1 ,then .

4) x =-1 ,then oscillates finitely.

5) x<-1 ,then oscillates infinitely.

And thus by using (1)

For -1<x<1 ,f(x) = 0 and

For x=1 , f(x)=1/2.

Thus D= (-1,1] = (-1,1)U{1}

Next we consider continuity over (-1,1].

Let -1<c<1 ,and consider .

HENCE f is continuous over (-1,1).

But if c=1 ,then f(1)=1/2 ,so we see there is a gap from 0 ,which is the value of f(x) for -1<x<1 , to 1/2 =f(1) and hence the function is not continuous.

To prove that:

Let, ε=1/4, then for all δ>0 choose ,x=0 ,then |f(x)-f(1)|=|f(0)-f(1)|=|0-1/2|= .

Thus .

Now to prove : if ,then ,where ,when x belongs to D.

For that you must consider the : .................................................. ..................................2

Now multiply (2) by and we get:

.

But ,thus :

.................................................. .......................................3

And if we want (3) smaller than ε>0 we must have :

which is possible since f is continuous at c.

Hence sqrt(f) is continuous at c