How do you use the formal definition of a limit to prove #lim(x^3 + 3x + 5) = 19# as x approaches 2?

Sketch of preliminary analysis

Proof

To use the formal definition of a limit to prove that lim(x^3 + 3x + 5) = 19 as x approaches 2, we need to show that for any positive value ε, there exists a positive value δ such that if 0 < |x - 2| < δ, then |(x^3 + 3x + 5) - 19| < ε.

Let's proceed with the proof:

Given ε > 0, we need to find a δ > 0 such that if 0 < |x - 2| < δ, then |(x^3 + 3x + 5) - 19| < ε.

First, let's simplify the expression |(x^3 + 3x + 5) - 19|:

|(x^3 + 3x + 5) - 19| = |x^3 + 3x - 14|

Now, we can proceed with the proof by finding an appropriate δ.

Let's consider the expression |x^3 + 3x - 14|:

|x^3 + 3x - 14| = |(x - 2)(x^2 + 2x + 7)|

Since we are interested in the behavior of the expression as x approaches 2, we can assume that 0 < |x - 2| < 1 (this assumption will be useful later).

Now, let's consider the expression |x^2 + 2x + 7|:

|x^2 + 2x + 7| = |(x + 1)^2 + 6|

Since (x + 1)^2 is always non-negative, we have:

|(x + 1)^2 + 6| ≥ 6

Now, let's choose δ = min{1, ε/6}. We will show that if 0 < |x - 2| < δ, then |(x^3 + 3x + 5) - 19| < ε.

Assume 0 < |x - 2| < δ. This implies that 0 < |x - 2| < 1 and 0 < |x - 2| < ε/6.

From the assumption 0 < |x - 2| < 1, we have:

|(x + 1)^2 + 6| ≥ 6

Now, let's consider the expression |(x - 2)(x^2 + 2x + 7)|:

|(x - 2)(x^2 + 2x + 7)| = |x - 2| |(x + 1)^2 + 6|

Since 0 < |x - 2| < ε/6, we have:

|x - 2| |(x + 1)^2 + 6| < (ε/6) * 6 = ε

Therefore, we have shown that if 0 < |x - 2| < δ (where δ = min{1, ε/6}), then |(x^3 + 3x + 5) - 19| < ε.

Hence, by the formal definition of a limit, we have proved that lim(x^3 + 3x + 5) = 19 as x approaches 2.