How do you prove that the limit #x^(1/2) = 4# as x approaches 16 using the formal definition of a limit?

ie where the function is continuous, the limit is the actual value of the function

To prove that the limit of x^(1/2) is equal to 4 as x approaches 16 using the formal definition of a limit, we need to show that for any positive value ε (epsilon), there exists a positive value δ (delta) such that if 0 < |x - 16| < δ, then |x^(1/2) - 4| < ε.

Let's proceed with the proof:

Given the function f(x) = x^(1/2), we want to prove that lim(x→16) f(x) = 4.

We start by assuming ε > 0. Our goal is to find a δ > 0 such that if 0 < |x - 16| < δ, then |x^(1/2) - 4| < ε.

Now, let's manipulate the expression |x^(1/2) - 4|:

|x^(1/2) - 4| = |√x - 4|

To simplify further, we can multiply the expression by the conjugate of the numerator:

|x^(1/2) - 4| = |√x - 4| * |√x + 4| / |√x + 4|

Expanding the numerator:

|x^(1/2) - 4| = |x - 16| / |√x + 4|

Since we want to find a δ such that |x^(1/2) - 4| < ε, we can rewrite the expression as:

|x - 16| / |√x + 4| < ε

Now, we can make an assumption that 0 < |x - 16| < 1, which implies that 15 < x < 17.

Using this assumption, we can find an upper bound for |√x + 4|:

|√x + 4| ≤ √17 + 4

Let's denote M = √17 + 4.

Now, we can rewrite the inequality as:

|x - 16| / M < ε

To ensure that the inequality holds, we can choose δ = ε * M.

Therefore, if 0 < |x - 16| < δ = ε * M, then |x^(1/2) - 4| < ε.

This completes the proof that the limit of x^(1/2) is equal to 4 as x approaches 16 using the formal definition of a limit.