Final answer:
To prove the limit lim[x→2] (x³ - 2x² - x + 3) = 1 using the δ-ε definition, we need to show that for any ε > 0, there exists a δ > 0 such that whenever 0 < |x - 2| < δ, then |(x³ - 2x² - x + 3) - 1| < ε.
Step-by-step explanation:
In order to prove the limit lim[x→2] (x³ - 2x² - x + 3) = 1 using the δ-ε definition, we need to show that for any ε > 0, there exists a δ > 0 such that whenever 0 < |x - 2| < δ, then |(x³ - 2x² - x + 3) - 1| < ε.
Let's start by simplifying the expression (x³ - 2x² - x + 3) - 1: (x³ - 2x² - x + 3) - 1 = x³ - 2x² - x + 2.
Now, we want to find a δ such that |x³ - 2x² - x + 2| < ε whenever 0 < |x - 2| < δ.
We can rewrite |x³ - 2x² - x + 2| as |(x - 2)(x² + x - 1)|, and since x² + x - 1 is a continuous function, it is bounded by some number M for x in the interval (1, 3).
So, let's choose M = max{1² + 1 - 1, 3² + 3 - 1} = max{1, 9} = 9. Now, to bound (x - 2), we can choose δ = min{1, ε/9}.
Therefore, whenever 0 < |x - 2| < δ, we have |(x³ - 2x² - x + 2) - 1| < ε, as required.