Prove that $f_n(x)$ converges uniformly on $[1,\infty)$ but not on $(0,1)$. Let $f_n(x) = \frac{\sin (nx)}{nx}$. Prove that $f_n(x)$ converges uniformly on $[1,\infty)$ but not on $(0,1)$.
Let $x\in [1,\infty)$. We observe that:
$$\frac{1}{nx} \le \frac{1}{n} \to 0$$
So by M-test, the sequence converges uniformly.
Next, we show that the sequence doesn't converge uniformly in the range $(0,1)$.
$$\lim_{n\to\infty} \sup_{x\in (0,1)} \left| \frac{\sin nx}{nx} \right| = \lim_{n\to\infty} \sup_{x\in (0,1)} \frac{1}{n} \left| \frac{\sin nx}{x} \right|$$
We notice that for every $n$, there's a sufficiently small $x$ such that:
$$\left| \frac{\sin nx}{x} \right| \ge n$$
So by choosing those $x_n$ we get the limit is $\ge 1 \ne 0$. Hence, there's no uniform convergence at $(0,1)$.
I'd like to get a review of my work (if it's right/rigorous)
Thanks.

You make at least two mistakes. First, Weierstrass's M-test concerns only series, while here you investigate sequences. Second, for $x \ge 0$ we always have $\sin x \le x$, which implies $\frac {\sin nx} x = \frac {\sin nx} {nx} n \le n$, so the inequality is in fact reversed (as opposed to how you have used it).
Now, the solution. First, note that on both intervals $\lim f_n = 0$ (pointwisely, not uniformly).
On $[1,\infty)$, making the change of variable $y=nx$, we get $0 \le \sup \limits _{x \ge 1} \frac {|\sin nx|} {nx} = \sup \limits _{y \ge n} \frac {|\sin y|} y \le \sup \limits _{y \ge n} \frac {|\sin y|} n \le \frac 1 n \to 0$, therefore $\sup \limits _{x \ge 1} \frac {|\sin nx|} {nx} \to 0$ so the convergence is uniform.
On $(0,1)$, note that $\sup \limits _{x \in (0,1)} \frac {|\sin nx|} {nx} \ge \frac {|\sin n \frac 1 n|} {n \frac 1 n} = \sin 1$, so it clearly cannot tend towards $0$, therefore the convergence is not uniform.
