Prove $f(x) = \frac{1}{x^2}$ is uniformly continuous on $[1, \infty]$ I am trying to prove this function is uniformily continuous on $[1, \infty]$, so far i have;
$$|f(x) - f(x)| = |\frac{1}{x^2} - \frac{1}{y^2}| = |\frac{(x-y)(x+y)}{x^2y^2}|$$
and then,
$$|x-y||\frac{x+y}{x^2y^2}|$$
I am not really sure where to go from this point?

Since $x, y\ge 1$ then
$$|x-y|\frac{x+y}{x^2y^2}\le |x-y|\frac{x^2+y^2}{x^2y^2}= |x-y|\left(\frac{1}{x^2}+\frac{1}{y^2}\right)$$
$$ =|x-y|\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\le 2|x-y|$$
