I would like to check whether $\sqrt{x} \sin(x)$ is uniformly continuous in its domain.
My attempt at question :
Since the given function is continuous everywhere therefore if the domain would have been bounded then the function would have been surely uniformly continuous , however since we are checking the uniform continuity in its whole domain therefore how should I approach the question ? Any help would be appreciated .
It is not uniformly continuous on $[0,\infty)$.
Take $x_n = 2n^2\pi$ and $y_n = 2n^2\pi + 1/n$. Then $|x_n - y_n| \to 0$ as $n \to \infty$ but
$$|f(x_n) - f(y_n)| = \sqrt{2n^2\pi +1/n}\sin(2n^2\pi+1/n) - \sqrt{2n^2\pi }\sin(2n^2\pi) \\ = n \sqrt{2\pi +1/n^3} \sin(1/n) \\ \to _{n \to \infty} \sqrt{2\pi} \neq 0.$$
Note also that the derivative is unbounded, but that is not sufficient for non-uniform continuity unless $f'(x) \to \infty$ -- which is not the case here.