Show that the function f (x) := 1/x2 is uniformly continuous on (1, ∞), but that it isnot uniformly continuous on (0, ∞)

Sure, let's break this down:

1. Uniform Continuity on (1, ∞):

A function f is uniformly continuous on an interval if for every ε > 0, there exists a δ > 0 such that for all x, y in the interval, if |x - y| < δ, then |f(x) - f(y)| < ε.

For the function f(x) = 1/x², let's take any ε > 0. We need to find a δ > 0 such that if x, y ∈ (1, ∞) and |x - y| < δ, then |f(x) - f(y)| < ε.

|f(x) - f(y)| = |1/x² - 1/y²| = |(y² - x²) / (x²y²)| = |(y - x)(y + x) / (x²y²)| ≤ |2δ / x²| (since |y - x| < δ and y + x < 2x for x, y ∈ (1, ∞)).

We want |2δ / x²| < ε, which is true if we choose δ = εx²/2. Since x > 1, δ can be any value less than ε/2. So, the function is uniformly continuous on (1, ∞).

2. Not Uniformly Continuous on (0, ∞):

To show that f is not uniformly continuous on (0, ∞), we need to find an ε > 0 such that for every δ > 0, there exist x, y ∈ (0, ∞) with |x - y| < δ but |f(x) - f(y)| ≥ ε.

Let's choose ε = 1. For any δ > 0, choose x = δ/2 and y = δ. Then |x - y| = δ/2 < δ, but |f(x) - f(y)| = |1/(δ/2)² - 1/δ²| = 4 - 1 = 3 ≥ ε.

So, the function f(x) = 1/x² is not uniformly continuous on (0, ∞).