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 ?
Hint: $\arcsin $ is a continuous function from $[-1,1]$ to $[0,2\pi]$. Continuous functions on a compact interval are uniformly continuous and composition of two uniformly continuous functions is uniformly continuous. Hence uniformly continuity of your function implies that of $x \sin x$. Can you show that $x \sin x$ is not uniformly continuous?