Question:

Show that the function: $\displaystyle f(x):=\frac{1}{x^2}$ is uniformly continous on $\displaystyle A:=[1,\infty)$, but that it is not uniformly continuous on $\displaystyle B := (0,\infty)$.

My thoughts:

So far I've gotten to $\displaystyle |f(x)-f(u)|\leq|x-u||x+u|$ when in A.