Prove $\frac{\sin x}{x^2}$ is uniformly continuous at $N(0;r)^c$ for any $r >0$ I tried $\left \vert \frac{\sin x}{x^2} -  \frac{\sin c}{c^2}\right \vert \leq \frac{1}{x^2} + \frac{1}{c^2} < \epsilon$, but it doesn't help me much with $\vert x - c \vert < \delta$. How can I prove this?

Note that for $x,y \neq 0$,
$$\left|\frac{\sin x}{x^2} - \frac{\sin y}{y^2} \right| = \left|\frac{\sin x}{x^2} - \frac{\sin x}{xy} + \frac{\sin x}{xy}- \frac{\sin y}{y^2}\right| \\\leqslant \left|\frac{\sin x }{x}\right|\left|\frac{1}{x} - \frac{1}{y}\right| + \frac{1}{|y|}\left| \frac{\sin x}{x} - \frac{\sin y}{y}\right|\\ = \left|\frac{\sin x }{x}\right|\left|\frac{1}{x} - \frac{1}{y}\right| + \frac{1}{|y|}\left| \frac{\sin x}{x} - \frac{\sin x }{y} + \frac{\sin x}{y}-\frac{\sin y}{y}\right|\\ \leqslant \left|\frac{\sin x }{x}\right|\left|\frac{1}{x} - \frac{1}{y}\right| + \frac{|\sin x|}{|y|}\left| \frac{1}{x} - \frac{1 }{y}\right|  + \frac{1}{|y|^2}\left|\sin x-\sin y\right|$$
Since $\left| \frac{\sin x}{x}\right|= \frac{|\sin x|}{|x|} \leqslant 1$ and $|\sin x | \leqslant 1$, it follows that
$$\left|\frac{\sin x}{x^2} - \frac{\sin y}{y^2} \right| \leqslant \left(1 + \frac{1}{|y|} \right)\left|\frac{1}{x} - \frac{1}{y}\right| +  \frac{1}{|y|^2}\left|\sin x-\sin y\right| \\ =\left(1 + \frac{1}{|y|} \right)\frac{|x-y|}{|x||y|} +  \frac{1}{|y|^2}\left|\sin x-\sin y\right|,$$
and for  all $x,y \in [r,\infty)\cup (-\infty,-r],$
$$\left|\frac{\sin x}{x^2} - \frac{\sin y}{y^2} \right| \leqslant \frac{1+r}{r^3}|x-y| + \frac{1}{r^2}|\sin x - \sin y|$$
Using the prosthaphaeresis formula , we get $|\sin x - \sin y | \leqslant |x - y|$ and it follows that
$$\left|\frac{\sin x}{x^2} - \frac{\sin y}{y^2} \right| \leqslant \frac{1+r}{r^3}|x-y| + \frac{1}{r^2}| x -  y| = \frac{1+2r}{r^3}|x-y|,$$
directly proving uniform continuity on $[r,\infty)\cup (-\infty,-r]$ where $r > 0$.
