Welcome to MSE! I need assistance in demonstrating the uniform convergence of the sequence $\frac{\sin nx}{n\sqrt{x}} \to 0$. I've attempted to approach the problem by trying a bounding technique, but my attempts have not been successful. Additionally, I've explored the possibility of finding a suitable sequence to show that it doesn't converge to zero, but my efforts in this direction have been unsuccessful as well.
Could someone help me understand where I might be going wrong in my attempts? Perhaps there's a different approach or insight that I'm missing. Your guidance and suggestions would be immensely valuable. Thank you!
Clearly, the sequence converges uniformly on any set that is bounded away from $0$. For example, if $x \in [a,\infty)$ where $a>0$, then uniform convergence holds since
$$\sup_{x \in [a,\infty)}\left|\frac{\sin nx}{n\sqrt{x}} \right|\leqslant \frac{1}{n\sqrt{a}}\underset{n \to \infty}\longrightarrow 0$$
To prove uniform convergence on all of $\mathbb{R}^+$ consider $x \in (0,a]$. If the sequence failed to converge uniformly then there would exist $\epsilon_0 > 0$ and a sequence $x_n \in (0,a]$ such that for all $n$
$$\left|\frac{\sin nx_n}{n\sqrt{x_n}} \right|\geqslant \epsilon_0$$
The sequence $(x_n)$ may or may not have a limit but it is bounded and $l=\liminf_{n \to \infty}x_n \geqslant 0.$
If $l =0$, then there is a subsequence $x_{n_k}$ such that $x_{n_k} \to 0$ and for all positive integrers $k$
$$\left|\frac{\sin n_kx_{n_k}}{n_k\sqrt{x_{n_k}}} \right|= \left|\frac{\sin n_kx_{n_k}}{n_kx_{n_k}} \right|\sqrt{x_{n_k}}\geqslant \epsilon_0$$
However, this is impossible because $\frac{\sin x}{x}$ is bounded for all $x \in \mathbb{R}$ and $\sqrt{x_{n_k}} \to 0$ as $ k \to \infty$.
On the other hand, if $l >0$, then $x_n > l/2$ eventually which means that there exists $N$ such that for all $n \geqslant N$ we have both
$$\left|\frac{\sin nx_{n}}{n\sqrt{x_{n}}} \right|\geqslant \epsilon_0,$$
and
$$\tag{*}\left|\frac{\sin nx_{n}}{n\sqrt{x_{n}}} \right| \leqslant \frac{1}{n\sqrt{l/2}}$$
which, again, is impossible since the RHS of (*) tends to $0$ as $n \to \infty$.
Using a proof by contradiction we have shown that the convergence of the sequence in question is uniform for all $x \in \mathbb{R}^+$.