Proving uniform continuity and uniform discontinuity

Question

Could someone please explain to me how to show uniform continuity and not uniformly continuous for the following:

$f(x) = \frac{1}{x^2}$ for $A = [1, \infty)$ show uniform continuity

$f(x) = \frac{1}{x^2}$ for $B = [0, \infty)$ show that $f$ is not uniformly continuous

Answer 1

1. $|\frac{1}{x^2}-\frac{1}{y^2}|\leq |\frac{y^2-x^2}{x^2y^2}|=|\frac{(y-x)(y+x)}{x^2y^2}|\leq |\frac{x+y}{x^2y^2}||y-x|\leq (|\frac{1}{xy^2}|+|\frac{1}{yx^2}|)|x-y|\leq 2|x-y|$

2.$\frac{1}{x^2}$ is not uniformly continuous. $|\frac{1}{x^2}-\frac{1}{y^2}|\leq |\frac{y^2-x^2}{x^2y^2}|$ Near $0$ quantity is very large.