Uniform convergence of $\frac{n^2\sin(x)}{1+n^2x}$ My goal is to show that
$$\frac{n^2\sin(x)}{1+n^2x}$$
does not converge uniformly on $S=(0,\infty)$ but does so on any compact subset of $S$. First, we find the limit function
$$\frac{n^2\sin(x)}{1+n^2x} \to \frac{\sin(x)}{x} \qquad \text{ as } n \to \infty$$
Now, to show that this doesn't uniformly converge on the whole interval I am really only interested in the boundary points (since any compact subset supposedly makes it uniformly convergent). As $x \to 0$ our function goes to $1$and as $x \to \infty$ our function goes to $0$. I am unsure of how to calculate
$$\left|\left|\frac{n^2\sin(x)}{1+n^2x}-\frac{\sin(x)}{x} \right|\right|_S$$ 
and, moreover, show that it is $0$. From there, on any compact subset of $S$ I am guessing that 
$$\left|\left|\frac{n^2\sin(x)}{1+n^2x}-\frac{\sin(x)}{x} \right|\right|=0$$ 
Will be zero since if we consider $[a,b] \subset S$ we have
$$\left|\left|\frac{n^2\sin(x)}{1+n^2x}-\frac{\sin(x)}{x} \right|\right| \leq \left|\frac{n^2\sin(a)}{1+n^2a}\right|+\left|\frac{\sin(a)}{a}\right|=0$$
Thanks for your help!

Be consider to following sequences x_n=1,x_m=mπ in this case this sequential function does not satisfy in Cauchy condition.lim n^2/1+n^2=1and lim n^2sinnπ/1+n^2=0 and we can choose  epsilon=1/2  |x_n-x_m|>1/2 so that it is not uniform convergence.
