Can I say $1/x^2$ is not uniformly continuous on $[0, \infty)$ because it is not defined when x = $0$?

Question

This is in my homework to show if $1/x^2$ is uniformly continuous on $[0, \infty)$. I'm thinking since the definition of uniform continuity says for all $ x, y \in $ the domain, $|f(x) - f(y)| < \epsilon$, I can say that it is not uniformly continuous since it is not defined on $0$. But I'm not very confident about my such argument.

Answer 1

Actually $1/{x^2}$ is not uniformly continuous on $(0,\infty)$, since at every point $x$ you need a different constant $\delta$ in the definition of continuity. As $x \to 0$, $\delta \to \infty$. You can visualize it by the slope of the derivative which decreases unbounded.

In my opinion it does not make sense to talk about $1/{x^2}$ on $[0, \infty)$, since this is not a function. (It needs to have a value at $0$).

Answer 2

Your argument is correct. For a function to be uniformly continuous it needs to be defined, which $1/x^2$ is not at $0$, and be continuous, which $1/x^2$ is not at $0$, and then some other things.

Usually the problem is posed to show $1/x^2$ is not uniformly continuous on $(0,\infty)$. Now your argument fails-$1/x^2$ is nicely continuous on $(0,\infty)$. It takes more work with the definition to show that it is not uniformly continuous. Now you need to show that if I give you an $\epsilon \gt 0$ and a $\delta \gt 0$ and claim that $|x-y| \lt \delta \implies |\frac 1{x^2}-\frac 1{y^2}| \lt \epsilon$ you can prove me wrong. It comes down to taking $x,y$ small enough.