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 $x$.