If I want to prove that the function$\ f(x)=\sin(x^2)$ is not uniformly continuous on the domain $\mathbb{R}$, I need to show that:
$\exists\epsilon>0$ $\forall\delta>0$ $\exists{x,y}\in\mathbb{R}\ : |x-y|<\delta$ and $|\sin(x^2) - \sin(y^2)|\geq\epsilon$.
So let's take $\epsilon = 1$. Then I want $|\sin(x^2)-\sin(y^2)|\ge1$. That's the case if $\sin(x^2)$=0 and $\sin(y^2)=\pm1$. Thus $x^2=n\pi$ and $y^2=n\pi + \frac{1}{2}\pi$. Now I'm stuck on expressing x and y, which I want to express in $\delta$, to ensure that $|x-y|<\delta$.
Thanks in advance for any help.