Use the ϵ, δ-definition of the limit of a function to prove lim x→2 x^2 + x + 1 = 7

To prove that \(\lim_{x \to 2} (x^2 + x + 1) = 7\) using the \(\epsilon\)-\(\delta\) definition of a limit, we must demonstrate that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x - 2| < \delta\), then \(|(x^2 + x + 1) - 7| < \epsilon\).

### Step-by-Step Proof:

#### 1. Restate the goal:
We want to show that for every \(\epsilon > 0\), we can find a \(\delta > 0\) such that:
\[
0 < |x - 2| < \delta \implies |(x^2 + x + 1) - 7| < \epsilon.
\]

#### 2. Simplify the expression \(|(x^2 + x + 1) - 7|\):
\[
|x^2 + x + 1 - 7| = |x^2 + x - 6|.
\]
Now factor \(x^2 + x - 6\):
\[
x^2 + x - 6 = (x - 2)(x + 3),
\]
so:
\[
|x^2 + x - 6| = |(x - 2)(x + 3)|.
\]

#### 3. Find a bound for \(|x + 3|\):
We need to control \(|(x - 2)(x + 3)|\). Notice that if \(x\) is close to 2, then \(|x + 3|\) will be close to 5 (since \(2 + 3 = 5\)). To make things easier, we assume \(\delta \leq 1\), which implies that \(|x - 2| < 1\), and thus \(1 < x < 3\). Therefore:
\[
4 < x + 3 < 6,
\]
so we have:
\[
|x + 3| < 6.
\]

#### 4. Bound \(|(x - 2)(x + 3)|\):
Now, using the bound \(|x + 3| < 6\), we get:
\[
|(x - 2)(x + 3)| = |x - 2| \cdot |x + 3| < 6|x - 2|.
\]

#### 5. Ensure \(|(x - 2)(x + 3)| < \epsilon\):
We want:
\[
6|x - 2| < \epsilon.
\]
This is equivalent to:
\[
|x - 2| < \frac{\epsilon}{6}.
\]

#### 6. Choose \(\delta\):
Thus, we can choose \(\delta = \min\left(1, \frac{\epsilon}{6}\right)\). This ensures that if \(0 < |x - 2| < \delta\), then:
\[
|(x - 2)(x + 3)| < \epsilon.
\]

#### 7. Conclusion:
By choosing \(\delta = \min\left(1, \frac{\epsilon}{6}\right)\), we have shown that for every \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(0 < |x - 2| < \delta\), then \(|(x^2 + x + 1) - 7| < \epsilon\). Therefore, by the \(\epsilon\)-\(\delta\) definition of a limit:
\[
\lim_{x \to 2} (x^2 + x + 1) = 7.
\]

Would you like further clarification or more details? 

Here are some related questions to deepen your understanding:
1. How does the \(\epsilon\)-\(\delta\) definition apply to more complex functions?
2. Can the same approach be used for limits at infinity?
3. How would the proof change if the function had a discontinuity at \(x = 2\)?
4. How do you interpret the geometrical meaning of the \(\epsilon\)-\(\delta\) definition?
5. How does factoring help simplify limit proofs in calculus?

**Tip:** When using the \(\epsilon\)-\(\delta\) definition, bounding certain terms (like \(|x + 3|\) here) can simplify finding \(\delta\). Always aim to reduce the number of terms that depend on \(x\).

Explore the epsilon-delta definition of limits with a detailed proof of lim x→2 (x^2 + x + 1) = 7. Learn step-by-step how to bound expressions and choose delta based on epsilon in this foundational calculus concept. Suitable for college-level calculus students.

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