$$f_n(x)=\frac{\sin(nx)}{\sqrt{n}}$$
Pointwise convergence:
$$\lim_{n \rightarrow \infty } \ f_n (x)=0$$
It converges to the function $f(x)\equiv0$
Uniform convergence
$$f'_n(x)=n \frac{\cos(nx)}{\sqrt{n}}$$ $$f''_n(x)=-n^2 \frac{\sin(nx)}{\sqrt{n}}$$
I have to find the points $x_0$ that: $$f'_n(x_0)=0 $$ and $$f''_n(x_0)<0 $$ to apply the definition of uniform convergence.
So: $$\cos(nx_0)=0 \Longleftrightarrow n x_0=\frac{\pi}{2}+k \pi$$ $$x_0=\frac{\pi}{2n}+\frac{k \pi}{n}$$
To respect $f''_n(x_0)<0 $:
$$x_0=\frac{\pi}{2n}+\frac{2k \pi}{n}$$
So:
$$f_n(x_0)=\frac{1}{\sqrt{n}}$$
and
$$\lim_{n \rightarrow \infty } \left| \frac{1}{\sqrt{n}}-0 \right|=0 $$
It respects uniform convergence
Is it correct?
Much easier: $$|f_n(x)-0| = \left|\frac{\sin(nx)}{\sqrt{n}}\right|\le\frac1{\sqrt{n}}.$$