Proof of $1/x^2$ being uniformly continuous on $[a,\infty)$ for $a>0$

Question

When I try to prove this I find that $$|f(x)-f(y)|=|\frac{1}{x^2}-\frac{1}{y^2}|<|\frac{1}{a^2}|+|\frac{1}{a^2}|=\frac{2}{a^2} $$ I can't see how to proceed by here, how can I get $|f(x)-f(y)|<\varepsilon$ given the above line?

Answer 1

To expand on the comment, $$\left|\frac 1 {y^2} - \frac 1 {x^2}\right|=\left| \frac {y^2-x^2}{x^2y^2}\right|=\left| \frac {(y-x)(y+x)}{x^2y^2}\right|=|y-x|\left| \frac {y+x}{x^2y^2}\right|$$

We can make $|y-x|$ small, so the key is that you now need to be able to bound $\left| \frac {y+x}{x^2y^2}\right|$ by a constant. Can you manage that?