Let $f_{n}(x) = \frac{n\sin(x)}{1+n^2\sin^2(x)}$ for $x \in [0,\pi]$. For $\delta > 0$, I want to find $E_{\delta} \subset [0, \pi]$ where $f_{n}(x)$ converges uniformly and $\mu([0,\pi] \setminus E_{\delta}) < \delta$ where $\mu$ denotes measure.
I've shown that $\lim_{n \to \infty} \sup_{x \in [0,\pi]} \left \{ \frac{n\sin(x)}{1+n^2\sin^2(x)} \right \} \neq 0$, but am not sure how to find this $E_{\delta}$. Ideas?
Let's look on the interval $[0,\pi/2]$ and choose an integer $m > 2$.
Then for $n > 2m^2$ and $x > \frac{1}{m}$, we have
\begin{eqnarray} 2m^2 < n &\iff& 2m < \frac{n}{m}\\ &\implies& 2m < nx \tag{since $x> \tfrac{1}{m}$}\\ &\implies& \frac{2}{nx} < \frac{1}{m} \end{eqnarray}
Now because $\sin (x) > \frac{x}{2}$ in $[0,\pi/2]$, we have $\frac{1}{\sin(x)} < \frac{2}{x}$ so that
$$\frac{n \sin(x)}{1+n^2 \sin^2(x)} < \frac{n \sin(x)}{n^2 \sin^2(x)} = \frac{1}{n \sin(x)} < \frac{1}{n \frac{x}{2}} = \frac{2}{nx} < \frac{1}{m}$$
Thus for large enough $n$, $f_n(x) < \frac{1}{m}$ on the interval $\left(\frac{1}{m},\frac{\pi}{2}\right)$.
You can apply a symmetry argument to get a similar result on the interval $[\pi/2,\pi]$.
Finally, for $\delta > 0$, pick $m > \frac{1}{3\delta}$ and $E_\delta = \left(\frac{1}{m},\pi-\frac{1}{m}\right)$