The sequence is: $$f_n(x) = \frac{n sin(x)}{1+n^2sin^2x}, x\in[0,\pi].$$
I believe it does uniformly converge and this is how I showed it.
$|f_n(x) - f(x)| = |\frac{n sin(x)}{1+n^2sin^2x} - 0| = |\frac{n sin(x)}{1+n^2sin^2x}|.$
Now since $sin(x)$ is bounded between 0 and 1 then $0\leq \frac{n sin(x)}{1+n^2sin^2x} \leq \frac{n}{1+n^2}$ where $\frac{n}{1+n^2} \rightarrow 0$ as $n \rightarrow \infty$
Is this correct? Can I show uniform convergence in this way? And if I am wrong how do I show if a sequence is or is not uniformly converging and for what set will this sequence converge uniformly.
Hint: $f_n \to 0$ uniformly on $[0,\pi]$ iff $\sup_{[0,\pi]} |f_n| \to 0.$ Consider $f_n(1/n).$