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, ∞).

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, we can choose δ = ε/2 to ensure that |f(x) - f(y)| < ε for all x, y ∈ (1, ∞) with |x - y| < δ. Therefore, f 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 take 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.

Therefore, f is not uniformly continuous on (0, ∞).

To show that the function f(x) = 1/x^2 is uniformly continuous on (1, ∞), we need to show that for any ε > 0, there exists a δ > 0 such that for all x, y in (1, ∞), if |x - y| < δ, then |f(x) - f(y)| < ε.

Step 1: Let's take ε > 0. We need to find δ > 0 such that if |x - y| < δ, then |f(x) - f(y)| < ε.

Step 2: The function f(x) = 1/x^2 is decreasing on (1, ∞). So, if x < y, then f(x) - f(y) = 1/x^2 - 1/y^2 = (y^2 - x^2) / (x^2 * y^2) = (y - x) * (y + x) / (x^2 * y^2).

Step 3: Since x, y are in (1, ∞), we have y + x < 2y and x^2 * y^2 > x^2. So, (y - x) * (y + x) / (x^2 * y^2) < (y - x) / x^2.

Step 4: We can make (y - x) / x^2 < ε by choosing δ = ε * x^2. So, if |x - y| < δ, then |f(x) - f(y)| < ε.

Therefore, the function f(x) = 1/x^2 is uniformly continuous on (1, ∞).

To show that the function f(x) = 1/x^2 is not uniformly continuous on (0, ∞), we need to show that there exists ε > 0 such that for any δ > 0, there exist x, y in (0, ∞) such that |x - y| < δ and |f(x) - f(y)| ≥ ε.

Step 1: Let's take ε = 1. We need to find δ > 0 and x, y in (0, ∞) such that |x - y| < δ and |f(x) - f(y)| ≥ 1.

Step 2: Let's take δ > 0 and choose x = δ/2 and y = δ. Then, |x - y| = δ/2 < δ.

Step 3: We have |f(x) - f(y)| = |1/(δ/2)^2 - 1/δ^2| = |4/δ^2 - 1/δ^2| = 3/δ^2.

Step 4: Since δ > 0, we have 3/δ^2 ≥ 1.

Therefore, the function f(x) = 1/x^2 is not uniformly continuous on (0, ∞).

Sure, let's break this down:

1. Uniform Continuity on (1, ∞):

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

For the function f(x) = 1/x², let's take any x, y in (1, ∞) and assume without loss of generality that x > y. Then:

|f(x) - f(y)| = |1/x² - 1/y²| = |y² - x²| / (x²y²) = |x - y| * |x + y| / (x²y²)

Since x, y are in (1, ∞), we have |x + y| / (x²y²) ≤ 2 (you can verify this by taking x = y = 1). So:

|f(x) - f(y)| ≤ 2|x - y|

This means that for every ε > 0, we can take δ = ε/2, and we have that if |x - y| < δ then |f(x) - f(y)| < ε. Therefore, f 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 in (0, ∞) with |x - y| < δ but |f(x) - f(y)| ≥ ε.

Let's take 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/x² - 1/y²| = |4 - 1| / (δ²/4) = 3 / (δ²/4) = 12/δ²

Since δ can be arbitrarily small, 12/δ² can be arbitrarily large, so it's always possible to find x, y such that |f(x) - f(y)| ≥ ε. Therefore, f is not uniformly continuous on (0, ∞).