Hello... i was trying to do this but stuck in between ...can anyone help me?

F (x)= xsin (1/x): x not equal to 0
F (x)= 0 : for x=0
For all x belongs closed interval -1,1..
Prove f (x) is uniformly continuous by using € defination?

Hello... i was trying to do this but stuck in between ...can anyone help me?

F (x)= xsin (1/x): x not equal to 0
F (x)= 0 : for x=0
For all x belongs closed interval -1,1..
Prove f (x) is uniformly continuous by using € defination?

After many years of lecturing on this material, I don't remember seeing a $\epsilon/\delta$ proof of this. What I have seen many times is the fact that $F(x)$ defined above is continuous on $[0,1]$ therefore it is uniformly continuous. If you find a direct proof, then I would like to see it.

Dear plato..i have seen its solution in maths forum but i am not satisfied with that..
There in € defination they had taken
|x1Sin (1/x1)-x2sin (1/x2)| is less than or equal to |x1sin (1/x1)-x2sin (1/x1)+x1sin (1/x2)-x2sin (1/x2)|
And then simplifying RHS we get 2 |x1-x2|...
They had proved in this way but i was thinking how middle both term is always greater than or equal to LHS .. N the reason they had given was that
Both extra added term are almost equal..how they can be equal?? Can you please help me?

Dear plato..i have seen its solution in maths forum but i am not satisfied with that..
There in € defination they had taken
|x1Sin (1/x1)-x2sin (1/x2)| is less than or equal to |x1sin (1/x1)-x2sin (1/x1)+x1sin (1/x2)-x2sin (1/x2)|

That's not correct- x2sin(1/x1) and x1 sin(1/x2) will not cancel. That last term should be |x1sin(1/x1)- x2sin(1/x1)+ x2sin(1/x1)- x2sin(1/x2)|= |(x1- x2)sin(1/x1)+ x2(sin(1/x1)- sin(1/x2))|.

And then simplifying RHS we get 2 |x1-x2|...
They had proved in this way but i was thinking how middle both term is always greater than or equal to LHS .. N the reason they had given was that
Both extra added term are almost equal..how they can be equal?? Can you please help me?