We are tasked to use the ε-δ definition of limits to prove:

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

The ε-δ Definition of Limits

For a function $f(x)$, the statement $\lim_{x \to a} f(x) = L$ means that for every $\varepsilon > 0$, there exists a $\delta > 0$ such that:

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

Here:

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

We need to find $\delta$ in terms of $\varepsilon$.

Step 1: Setup the condition $|f(x) - L| < \varepsilon$

Start with:

$|f(x) - 1| < \varepsilon.$

Substitute $f(x) = x^2 + x - 2$ and $L = 1$:

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

Simplify the expression inside the absolute value:

$|x^2 + x - 3| < \varepsilon.$

Step 2: Factor the expression $x^2 + x - 3$

To simplify further, rewrite $x^2 + x - 3$ by factoring or approximating near $x = 2$.

1. Start from $x^2 + x - 3$: $x^2 + x - 3 = (x - 2)(x + 3).$

Thus, the inequality becomes:

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

Step 3: Control $|x + 3|$ using $\delta$

To isolate $|x - 2|$, note that $x$ is close to 2, so $x + 3$ will be close to 5 (since $2 + 3 = 5$).

We can restrict $|x - 2| < 1$ (this choice is arbitrary but simplifies calculations). Under this condition:

$x \text{ is in } (2 - 1, 2 + 1), \text{ or } x \in (1, 3).$

Thus:

$x + 3 \text{ is in } (1 + 3, 3 + 3), \text{ or } x + 3 \in (4, 6).$

This gives:

$|x + 3| \leq 6 \quad \text{(since the maximum value in this interval is 6).}$

Step 4: Relate $\varepsilon$ and $\delta$

Now we have:

$|(x - 2)(x + 3)| < \varepsilon \quad \text{and} \quad |x + 3| \leq 6.$

Therefore:

$|x - 2| \cdot |x + 3| < \varepsilon \implies |x - 2| \cdot 6 < \varepsilon.$

Divide both sides by 6 to isolate $|x - 2|$:

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

Step 5: Finalize $\delta$

To satisfy the definition, choose:

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

Step 6: Verification

If $0 < |x - 2| < \delta$, then:

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

2. $|x - 2| < 1$ ensures $|x + 3| \leq 6$.

Thus:

$|f(x) - 1| = |(x - 2)(x + 3)| < \varepsilon.$

Conclusion

We have shown that for any $\varepsilon > 0$, choosing $\delta = \min\left(1, \frac{\varepsilon}{6}\right)$ guarantees $|f(x) - 1| < \varepsilon$ whenever $0 < |x - 2| < \delta$. Therefore:

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

Do you want more clarification on any step? Here are 5 related questions to practice or expand on:

1. Prove $\lim_{x \to 1} (x^2 - 1) = 0$ using the $\varepsilon-\delta$ definition.
2. How does the $\varepsilon-\delta$ definition of a limit differ for linear versus quadratic functions?
3. Verify the limit $\lim_{x \to 3} (x^2 - 9) = 0$ step-by-step.
4. Why is the restriction $|x - 2| < 1$ helpful in this proof?
5. Prove $\lim_{x \to a} (x^2) = a^2$ using the $\varepsilon-\delta$ definition.

Tip: When working with quadratic functions, factorizing the expression and controlling the terms simplifies the proof significantly!