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?
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?