We are tasked with finding a number $\delta$ such that the following inequality holds:

$|x^2 - 1| < 0.2 \quad \text{whenever} \quad 0 < |x - 1| < \delta.$

This is essentially a problem of finding the appropriate $\delta$ to satisfy the given conditions based on the behavior of the function $f(x) = x^2$.

Step-by-Step Solution:

1. Rewrite the condition: We need to find $\delta$ such that if $|x - 1| < \delta$, then $|x^2 - 1| < 0.2$. This can be interpreted as an $\varepsilon-\delta$ definition of continuity at $x = 1$, with $\varepsilon = 0.2$.

2. Simplify $|x^2 - 1|$: Notice that: $|x^2 - 1| = |(x - 1)(x + 1)|.$ So the inequality becomes: $|(x - 1)(x + 1)| < 0.2.$

3. Bound $|x + 1|$: Since we are interested in $x$ values close to $1$, we can make an estimate. If $|x - 1| < \delta$, then $x$ is close to $1$. Specifically, we know that $x$ will be in the range $(1 - \delta, 1 + \delta)$, so: $|x + 1| \approx 2 \quad \text{when} \quad x \text{ is near } 1.$ This is a rough approximation for $x$ values close to $1$.

4. Solve for $\delta$: Using the approximation $|x + 1| \approx 2$, the inequality $|(x - 1)(x + 1)| < 0.2$ becomes: $|x - 1| \cdot 2 < 0.2,$ which simplifies to: $|x - 1| < \frac{0.2}{2} = 0.1.$

Thus, we find that $\delta = 0.1$ should work. That is, if $|x - 1| < 0.1$, then $|x^2 - 1| < 0.2$.

Conclusion:

The number $\delta$ that satisfies the condition is $\delta = 0.1$.

Would you like more details, or do you have any questions? Here are some follow-up questions:

1. How can the $\varepsilon-\delta$ definition of continuity be applied to this problem?
2. What would happen if $\varepsilon = 0.1$ instead of $0.2$?
3. Why is it necessary to approximate $|x + 1|$ when solving for $\delta$?
4. Can the value of $\delta$ be larger than $0.1$ for some other choice of approximation?
5. How does the graph of $f(x) = x^2$ help visualize the solution?

Tip: When solving $\varepsilon-\delta$ problems, a good first step is often to rewrite expressions in factored form to make the relationship between variables clearer.