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!
For $0 &lt; a \leqslant x$ we have

$$\left|\frac{n^2\sin(x)}{1+n^2x}-\frac{\sin(x)}{x} \right| = \left|\frac{\sin (x)}{n^2x^2 + x}\right| \leqslant \frac{1}{n^2a^2 + a} \to 0
$$

and convergence is uniform on $[a, \infty)$.

For $x &gt; 0$

we have

$$\left|\frac{n^2\sin(x)}{1+n^2x}-\frac{\sin(x)}{x} \right| = \left|\frac{\sin (x) /x}{n^2x + 1}\right|
$$

Choose a sequence $x_n = 1/n^2$ to show that the convergence is not uniform on $(0,\infty)$.

Here we have $\sin (x_n) /x_n \geqslant \sin 1$ and

$$\left|\frac{n^2\sin(x_n)}{1+n^2x_n}-\frac{\sin(x_n)}{x_n} \right| \geqslant \frac{\sin 1}{2} $$
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.