uniform convergence of the sine function Find if $\sin  \frac{nx}{n+x}$ converges uniformly in $[0, K]$ and $[K, \infty)$
I'm a bit stuck, I tried writing
$\sup|\sin\frac{nx}{n+x}-\sin x|=\sup|\sin\frac{2nx+x^2}{2(n+x)}\cos\frac{x^2}{2(n+x)}|$
using the diference formula and finding a derivative, but it seems to be too complicated to be useful. Surely there is a more elegant solution.
Thanks for any help.

On $[0,K]$ use the mean value theorem to get
$$
\Bigl|\sin\frac{n\,x}{n+x}-\sin x\Bigr|\le\Bigl|\frac{n\,x}{n+x}-x\Bigr|=\frac{x^2}{n+x}\le\frac{K^2}{n-K}
$$
for all $n>K$. The convergence is thus uniform.
On $[K,\infty)$ choose $x=n\,\pi$ ($n>K$):
$$
\Bigl|\sin\frac{n\,x}{n+x}-\sin x\Bigr|_{x=n\pi}=\Bigl|\sin\frac{n\,\pi}{1+\pi}\Bigr|,
$$
that does not converge to $0$. The convergence is not uniform.
