Given, $f (x) = \sin (x \sin x)$ and we have to check if the function is uniformly continuous on $(0, \infty)$. So, far I haven't been successful but I tried to use this theorem " A function $f(x)$ is uniformly continuous on $(a,b)$ iff the extension of the function $f$ is continuous on $[a,b]$ "
Defining $f (x) = \sin (x \sin x)$, for $x$ in $(0, \infty)$ and $f (0) = 0$ I tried to show it continuous on $[0, \infty)$. But I think my procedure isn't quite right and the theorem could only be used for intervals of finite length.
Any suggestion regarding how to show it isnt uniformly continuous ?