# Math Help - uniformly continuous

1. ## uniformly continuous

How do i prove that x*sin(x) is not uniformly continuous?

2. Originally Posted by math8553
How do i prove that x*sin(x) is not uniformly continuous?
$x \sin(x)-(x+\delta) \sin(x+\delta)=x \delta \cos(x) + xO(\delta^2)$

and so for $\delta$ small enough and $x \ne n \pi/2, \ n\in \mathbb{Z}$ there is a $k>0$ independent of $x$ such that:

$x \sin(x)-(x+\delta) \sin(x+\delta)

So however small we make $|\delta|>0$ there is an $x$ which makes $|x \sin(x)-(x+\delta) \sin(x+\delta)|$ as large as we like.

Note: The $O(\delta^2)$ term can be made independent of $x$ as it stands in place of a remainder term of a Taylor series which is the product of a constant $\kappa$, $\delta^2$ and a circular trig function at a point in $[x,x+\delta]$ all of which which is bounded by $|\kappa \delta^2|$. If the reader needs the detail they can develop it them selves.

CB

3. Originally Posted by math8553
How do i prove that x*sin(x) is not uniformly continuous?
I think CB assumed you meant uniformly continuous on $\mathbb{R}$. You need to specify these things. This function is actually uniformly continuous on any bounded subspace of $\mathbb{R}$

4. Originally Posted by Drexel28
I think CB assumed you meant uniformly continuous on $\mathbb{R}$. You need to specify these things. This function is actually uniformly continuous on any bounded subspace of $\mathbb{R}$
That is what i assumed and for the reason you give.

(I had expected some complaints rather than thanks for my demonstration as it is not a conventional way of doing these things, with the real detail left to the reader to confirm).

CB

5. Originally Posted by CaptainBlack
T
with the real detail left to the reader to confirm).
Yes, but that is the way it should be.

Plus. New and strange methods are my favorite!