How to prove that $\sin(x \sin x)$ is not uniformly continuous on $\Bbb{R}^+$?
My attempt: I know that $\sin x$ is uniformly continuous on that interval when $\sin{x^2}$ is not.
Yes It is duplicate of one question. But the answer was not satisfactory. That's why I am posting it again.
Any help will be highly appreciated.
Hint :
Take, $x_n=\{2n\pi \}_n$ and $y_n=\{2n\pi +\frac{\pi}{n}\}_n$. Then $|x_n-y_n|\to 0$ as $n \to \infty$.
But , $|f(x_n)-f(y_n)|\to \sin(2\pi^2)\not= 0$
Edit :
$$\lim_{n\to \infty}\left(2n\pi+\frac{\pi}{n}\right)\sin\left(2n\pi+\frac{\pi}{n}\right)=\lim_{n\to \infty}\left(2n\pi+\frac{\pi}{n}\right)\sin(\pi /n)=\lim_{n\to \infty}2n\pi \sin(\pi /n)=2\pi ^2$$ So , $$\lim_{n\to \infty}\sin \left[\left(2n\pi+\frac{\pi}{n}\right)\sin\left(2n\pi+\frac{\pi}{n}\right)\right]=\sin \left[\lim_{n\to \infty}\left(2n\pi+\frac{\pi}{n}\right)\sin\left(2n\pi+\frac{\pi}{n}\right)\right]=\sin(2.\pi^2)$$