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It looks like you're working on a problem involving the limit definition and finding a delta (\(\delta\)) for a given epsilon (\(\epsilon\)) for the function \(f(x) = x^2\). This appears to be a classic epsilon-delta proof setup.

### Problem Setup:
The question is asking to find a number \(\delta\) such that for any \(\epsilon = \frac{1}{2}\), if \(|x - 1| < \delta\), then \(|x^2 - 1| < \frac{1}{2}\).

#### Steps to solve:
1. Start with the given inequality \(|x^2 - 1| < \frac{1}{2}\).
   
2. Factor \(|x^2 - 1| = |(x - 1)(x + 1)|\).

3. This inequality becomes \(|(x - 1)(x + 1)| < \frac{1}{2}\).

4. Since we are trying to bound \(|x - 1|\), we need a bound for \(|x + 1|\). We typically assume that \(|x - 1| < 1\), which implies \(0 < x < 2\), so \(|x + 1| < 3\).

5. Now, we have \(|x - 1||x + 1| < \frac{1}{2}\), and since \(|x + 1| < 3\), we get \(|x - 1| \cdot 3 < \frac{1}{2}\), or \(|x - 1| < \frac{1}{6}\).

Thus, for \(\epsilon = \frac{1}{2}\), we choose \(\delta = \frac{1}{6}\).

Would you like further details on any part of this explanation or clarification on the steps?

### Related Questions:
1. How does the choice of \(|x + 1| < 3\) affect the \(\delta\) calculation?
2. What happens if we choose a different \(\epsilon\)?
3. Can you explain how the epsilon-delta definition applies to other functions, such as \(f(x) = \sin(x)\)?
4. What role does factoring play in simplifying the inequality for this limit?
5. How would the solution change if the limit point was something other than \(x = 1\)?

**Tip:** When working on epsilon-delta proofs, it often helps to first estimate the range of \(x\) based on the initial assumption \(|x - a| < \delta\).

Use the given graph of f(x) = x^2 to find a number delta such that if |x - 1| < delta, then |x^2 - 1| < 1/2.

Learn how to find delta for an epsilon-delta proof involving the quadratic function f(x) = x^2. The solution demonstrates factoring the quadratic expression and bounding delta for a given epsilon. Suitable for students learning about limits in calculus.

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