# Uniform Continuity of 1/x^2 on various sets

1. Nov 21, 2008

### Lazerlike42

1. The problem statement, all variables and given/known data

Show that f(x)=$$\frac{1}{x^{2}}$$ is uniformly continuous on the set [1,$$\infty$$) but not on the set (0,1].

2. Relevant equations

3. The attempt at a solution

I've been working at this for at least 2 hours now, possibly 3, and I can't say I really have much of any idea about it. I've primarily been looking at the fact that a function f: A$$\rightarrow$$ $$\Re$$ is not uniformly continuous iff $$\exists$$ some $$\epsilon_{0}$$ > 0 and sequences xn and yn where |xn-yn| $$\rightarrow$$ 0 but |f(xn)-f(yn)|$$\geq$$ $$\epsilon$$0.

My thought was to somehow say that, for part one, because all sequences must be $$\geq$$ 1, they must converge to the same thing and therefore |f(xn)-f(yn)|= 0 for some n, but I'm not sure if I can for various reasons. In fact, I can't even really come up with any sequences in (0,1] to show that the function is not uniformly continuous on that set. I can think of another way to do that one, but I really want to see some sequences to do it to prove that it works in that case.

Any help is greatly appreciated.

2. Nov 21, 2008

### Staff: Mentor

Look at the graph of f(x) = 1/x^2. For the interval x >= 1, the graph is fairly level, so if |x1 - x2| is small, |f(x1) = f(x2)| is small as well. OTOH, the graph is very steep for x close to zero, so |x1 - x2| being small doesn't guarantee that |f(x1) - f(x2)| will also be small.