To formally prove that [tex]\(\lim_{x \rightarrow 2} \left(x^3 - 2x^2 + 4x\right) = 8\)[/tex], we need to show that for any [tex]\(\varepsilon > 0\)[/tex], there exists a [tex]\(\delta > 0\)[/tex] such that whenever [tex]\(0 < |x - 2| < \delta\)[/tex], it follows that [tex]\(|(x^3 - 2x^2 + 4x) - 8| < \varepsilon\)[/tex].
Firstly, define the function:
[tex]\[ f(x) = x^3 - 2x^2 + 4x \][/tex]
We are interested in the behavior of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 2. Let's calculate:
[tex]\[ f(2) = 2^3 - 2 \cdot 2^2 + 4 \cdot 2 = 8 - 8 + 8 = 8 \][/tex]
Now we consider:
[tex]\[ |f(x) - 8| = |x^3 - 2x^2 + 4x - 8| \][/tex]
We need this expression to be less than [tex]\(\varepsilon\)[/tex] for [tex]\(0 < |x - 2| < \delta\)[/tex].
Simplify the expression [tex]\(|x^3 - 2x^2 + 4x - 8|\)[/tex]:
[tex]\[ f(x) - 8 = x^3 - 2x^2 + 4x - 8 \][/tex]
Next, let's factor [tex]\((x - 2)\)[/tex] out of the polynomial:
[tex]\[
x^3 - 2x^2 + 4x - 8 = (x - 2)(x^2 + x + 4)
\][/tex]
So we have:
[tex]\[
|f(x) - 8| = |(x - 2)(x^2 + x + 4)|
\][/tex]
We need to find an [tex]\(m\)[/tex] such that:
[tex]\[
|(x - 2)(x^2 + x + 4)| < m| x - 2|
\][/tex]
This simplifies to ensuring:
[tex]\[
|x^2 + x + 4| < m
\][/tex]
We need to bound [tex]\(x^2 + x + 4\)[/tex] near [tex]\(x = 2\)[/tex]. Note that as [tex]\(x \to 2\)[/tex],
[tex]\[
x^2 + x + 4 \to 2^2 + 2 + 4 = 4 + 2 + 4 = 10
\][/tex]
Hence, we need to find [tex]\(m\)[/tex] such that:
[tex]\[
|x^2 + x + 4| \leq 10 \text{ within the chosen } \delta \text{ neighborhood }
\][/tex]
Therefore, we choose:
[tex]\(
m = 10
\)[/tex]
Finally, to ensure that our [tex]\(\delta\)[/tex] works correctly, we take:
[tex]\[
\delta = \min \left( \frac{\varepsilon}{m}, 1 \right) = \min \left( \frac{\varepsilon}{10}, 1 \right)
\][/tex]
Thus, the value of [tex]\(m\)[/tex] ensuring the condition for our δ-ε proof is:
[tex]\[
m = 10
\][/tex]
