For the limit

lim x → 3 (x3 − 4x + 4) = 19
illustrate the definition by finding the largest possible values of δ that correspond to ε = 0.2 and ε = 0.1. (Round your answers to four decimal places.)

To illustrate the definition and find the corresponding values of δ, we need to determine the largest possible values of δ that yield a limit within the given epsilon (ε) values.

Let's start by rewriting the expression and understanding the definition:

lim x → 3 (x^3 − 4x + 4) = 19

The definition states that for every epsilon (ε) greater than zero, there exists a delta (δ) greater than zero such that if the distance between x and the limit point (3 in this case) is less than delta (|x - 3| < δ), then the distance between the function value (f(x)) and the limit (19) is less than epsilon (|f(x) - 19| < ε).

We have ε = 0.2 and ε = 0.1. We will solve for δ in each case.

For ε = 0.2:
|f(x) - 19| < 0.2

Now, let's manipulate the expression to isolate x:

|f(x) - 19| = |x^3 − 4x + 4 - 19|
= |x^3 − 4x - 15|

To find the values of x that satisfy the inequality, we need to isolate x:

|x^3 − 4x - 15| < 0.2

One way to solve this is by graphing the function and finding the range of x values that satisfy the inequality. Another way is to use numerical methods such as Newton's method or bisection method to iteratively approximate the roots.

For ε = 0.1, we follow the same steps but substitute ε with 0.1 in the equations.

After solving for x, we can find δ by taking the maximum difference between 3 and the x values we found that satisfy the inequalities. Round the values of δ to four decimal places.

Note: The process for finding δ depends on the specific problem and the function. It may involve different methods, such as algebraic manipulation, graphical analysis, or numerical techniques.

To find the largest possible values of δ that correspond to ε = 0.2 and ε = 0.1, we can use the definition of a limit.

According to the definition, for any ε > 0, there exists a δ > 0 such that |f(x) - L| < ε whenever 0 < |x - c| < δ. In this case, f(x) = x^3 - 4x + 4 and L = 19, and c = 3.

Let's start with ε = 0.2:

|f(x) - 19| < 0.2

Now, let's simplify the expression:

|x^3 - 4x + 4 - 19| < 0.2

|x^3 - 4x - 15| < 0.2

Factoring the expression:

|(x - 3)(x^2 + 3x + 5)| < 0.2

We want to find the largest possible value of δ for which this inequality holds. To do that, we can find the maximum value of the expression inside the absolute value:

x^2 + 3x + 5 ≤ k, where k is a constant.

Now, we want to find the value of k such that the inequality holds for all x satisfying |x - 3| < δ.

To find the maximum value of x^2 + 3x + 5, we can complete the square:

x^2 + 3x + 5 = (x + 3/2)^2 + 5 - 9/4

So, x^2 + 3x + 5 ≤ (x + 3/2)^2 + 11/4

Now, we set the right side of the inequality equal to ε:

(x + 3/2)^2 + 11/4 ≤ ε

(x + 3/2)^2 ≤ ε - 11/4

x + 3/2 ≤ √(ε - 11/4)

x ≤ √(ε - 11/4) - 3/2

Now, we can set up the inequality:

δ = min{1, √(ε - 11/4) - 3/2}

Now, let's substitute ε = 0.2 into the equation to find the largest possible value of δ for this case:

δ = min{1, √(0.2 - 11/4) - 3/2}

δ = min{1, √(0.2 - 2.75) - 1.5}

δ = min{1, √(-2.55) - 1.5}

Since the square root of a negative number is undefined, δ is equal to 1.

Similarly, we can find the largest possible value of δ for ε = 0.1:

δ = min{1, √(0.1 - 11/4) - 3/2}

δ = min{1, √(-2.65) - 1.5}

Again, the square root of a negative number is undefined, so δ is equal to 1.

Therefore, the largest possible values of δ that correspond to ε = 0.2 and ε = 0.1 are both equal to 1.