Prove the theorem that if limx¡úd h(x)= P and limx¡úd k(x) = Q and h(x) ¡Ý k(x) for all x in an open interval containing d, then P ¡Ý Q by using the formal definition of the limit, showing all work.

To prove the theorem using the formal definition of the limit, we will need to start by stating what it means for the limit of a function to equal a specific value.

Formal Definition of a Limit: Let f(x) be a function defined on an open interval containing a point x = c, except possibly at c itself. We say that the limit of f(x) as x approaches c is L, denoted as lim┬(x→c)⁡〖f(x) = L 〗, if for every positive number ε, there exists a positive number δ such that if 0 < |x - c| < δ, then |f(x) - L| < ε.

Now, let's proceed with the proof:

Given: lim┬(x→d)⁡h(x) = P (1)
lim┬(x→d)⁡k(x) = Q (2)
h(x) ≥ k(x) for all x in an open interval containing d

To prove: P ≥ Q

Proof:
Using the formal definition of the limit, we know that for the given limits (1) and (2), we can choose positive values ε₁ and ε₂, and find corresponding values δ₁ and δ₂ such that:
1. For lim┬(x→d)⁡h(x) = P:
- For any positive ε₁, there exists a positive δ₁ such that if 0 < |x - d| < δ₁, then |h(x) - P| < ε₁.
- This gives us the inequality: P - ε₁ < h(x) < P + ε₁. (3)

2. For lim┬(x→d)⁡k(x) = Q:
- For any positive ε₂, there exists a positive δ₂ such that if 0 < |x - d| < δ₂, then |k(x) - Q| < ε₂.
- This gives us the inequality: Q - ε₂ < k(x) < Q + ε₂. (4)

Now, notice that we have h(x) ≥ k(x) for all x in an open interval containing d. Combining this inequality with (3) and (4), we get:
P - ε₁ < h(x) < P + ε₁
Q - ε₂ < k(x) < Q + ε₂

Since h(x) is greater than or equal to k(x), we can conclude:
P - ε₁ < h(x) ≥ k(x) > Q - ε₂

Add ε₂ to both sides of the inequality:
P - ε₁ + ε₂ < h(x) + ε₂ ≥ k(x) > Q - ε₂ + ε₂
P - ε₁ + ε₂ < h(x) + ε₂ ≥ k(x) > Q

As ε₁ and ε₂ are positive values, we can rewrite the inequality as:
P - (ε₁ - ε₂) < h(x) + ε₂ ≥ k(x) > Q

Now, let's consider the quantity h(x) + ε₂. Since this quantity is greater than or equal to h(x), we have:
h(x) ≤ h(x) + ε₂

Combining the above inequality with the previous inequality, we get:
P - (ε₁ - ε₂) < h(x) + ε₂ ≥ k(x) > Q
P - (ε₁ - ε₂) < h(x) ≤ h(x) + ε₂ ≥ k(x) > Q

Now, notice that as x approaches d, the distance |x - d| will become smaller. We can choose ε₁ - ε₂ to be a positive value, say δ. Then, we have:
0 < |x - d| < δ

Applying this to the previous inequality, we get:
P - δ < h(x) ≤ h(x) + ε₂ ≥ k(x) > Q

Since h(x) + ε₂ is greater than or equal to k(x), we can conclude:
P - δ < h(x) ≥ k(x) > Q

This inequality holds for all x in an open interval containing d. Taking the limit as x approaches d, we have:
P - δ ≤ lim┬(x→d)⁡h(x) ≥ lim┬(x→d)⁡k(x) > Q

From (1) and (2):
P ≤ P ≥ Q

By transitivity, we can conclude that P ≥ Q.

Therefore, we have proven that if lim┬(x→d)⁡h(x) = P, lim┬(x→d)⁡k(x) = Q, and h(x) ≥ k(x) for all x in an open interval containing d, then P ≥ Q.