To solve this problem using the ε-δ definition of a limit, we need to prove the following statement rigorously:

$\lim_{x \to 2} (x^2 + x - 2) = 1.$

The ε-δ Definition of a Limit

For any function $f(x)$, we say:

$\lim_{x \to c} f(x) = L$

if for every $\varepsilon > 0$, there exists a $\delta > 0$ such that:

$0 < |x - c| < \delta \implies |f(x) - L| < \varepsilon.$

In this case:

• $f(x) = x^2 + x - 2$,
• $c = 2$,
• $L = 1$.

Step 1: Restate the Condition to Prove

We need to show that for every $\varepsilon > 0$, there exists a $\delta > 0$ such that:

$0 < |x - 2| < \delta \implies |(x^2 + x - 2) - 1| < \varepsilon.$

Simplify the expression inside the absolute value:

$|(x^2 + x - 2) - 1| = |x^2 + x - 3|.$

Thus, we need:

$|x^2 + x - 3| < \varepsilon \quad \text{whenever} \quad 0 < |x - 2| < \delta.$

Step 2: Factorize and Simplify

We now focus on $|x^2 + x - 3|$. Start by factoring the quadratic expression:

$x^2 + x - 3 = (x - 2)(x + 3).$

Thus, the condition becomes:

$|(x - 2)(x + 3)| < \varepsilon.$

Step 3: Bounding $|x + 3|$

To control $|(x - 2)(x + 3)|$, we need to bound $|x + 3|$. If $x$ is close to 2, say $|x - 2| < 1$, then:

$|x - 2| < 1 \implies 1 < x < 3.$

From this, we can estimate $|x + 3|$:

$1 < x < 3 \implies 4 < x + 3 < 6 \quad \implies \quad |x + 3| < 6.$

Step 4: Choose $\delta$

Now we return to the condition $|(x - 2)(x + 3)| < \varepsilon$. Since $|x + 3| < 6$, we have:

$|(x - 2)(x + 3)| < |x - 2| \cdot 6.$

To ensure $|(x - 2)(x + 3)| < \varepsilon$, we can set:

$|x - 2| \cdot 6 < \varepsilon.$

Dividing both sides by 6:

$|x - 2| < \frac{\varepsilon}{6}.$

Thus, we can choose:

$\delta = \min\left(1, \frac{\varepsilon}{6}\right).$

Step 5: Verify the Solution

Now, let $0 < |x - 2| < \delta$, where $\delta = \min\left(1, \frac{\varepsilon}{6}\right)$. Then:

1. Since $|x - 2| < 1$, we know $|x + 3| < 6$.
2. Therefore:

$|(x - 2)(x + 3)| < |x - 2| \cdot 6.$

3. Since $|x - 2| < \frac{\varepsilon}{6}$, it follows that:

$|(x - 2)(x + 3)| < \frac{\varepsilon}{6} \cdot 6 = \varepsilon.$

Thus, we have shown that:

$0 < |x - 2| < \delta \implies |(x^2 + x - 2) - 1| < \varepsilon.$

Conclusion

By the ε-δ definition of a limit, we have rigorously proven that:

$\lim_{x \to 2} (x^2 + x - 2) = 1.$

Do you want me to elaborate further on any part? Let me know!

Here are 5 related questions to expand your understanding:

1. What is the ε-δ definition of a limit and why is it important?
2. How do you prove the limit of a linear function using the ε-δ definition?
3. Can you explain why bounding $|x + 3|$ is critical in this proof?
4. How would the proof change if the function were $x^2 - 4$?
5. How can you visually interpret the ε-δ definition on a graph?

Tip: In ε-δ proofs, always simplify and factor the function as much as possible to identify key terms you need to control.