Find a positive number $\delta$? Find a positive number $\delta$ such that when $|x-1|<\delta$, then $|x^2-1| < 0.45$
Part 2:
Find the LARGEST number $\lambda$ with the property that when $|x-1|< \lambda$, then $|x^2-1| < 0.45$
I don't understand this at all. Help?

Suppose $0 < \epsilon < 1$. Consider the case $x^2 - 1 \ge 0$ first, i.e., $x \in (-\infty, -1] \cup [1, \infty)$. In this case, the inequality $|x^2 - 1| < \epsilon$ becomes
$$
x^2 - 1 - \epsilon < 0.
$$
Factor this as
$$
(x - \sqrt{1 + \epsilon})(x + \sqrt{1 + \epsilon}) < 0.
$$
This is true when $x \in (-\sqrt{1 + \epsilon}, \sqrt{1 + \epsilon})$. So we get $x \in (-\sqrt{1 + \epsilon}, -1] \cup [1, \sqrt{1 + \epsilon})$ for this case.
The other case is when $x^2 - 1 \le 0$, i.e., $x \in [-1, 1]$, the inequality $|x^2 - 1| < \epsilon$ becomes
$$
x^2 - 1 + \epsilon > 0.
$$
Factor this as
$$
(x - \sqrt{1 - \epsilon})(x + \sqrt{1 - \epsilon}) > 0.
$$
This is true when $x \in (-\infty, -\sqrt{1 - \epsilon}) \cup (\sqrt{1 - \epsilon}, \infty)$, so we get $x \in [-1, -\sqrt{1 - \epsilon}) \cup (\sqrt{1 - \epsilon}, 1]$ for this case. Combining them both, we get
$$
x \in (-\sqrt{1 + \epsilon}, -\sqrt{1 - \epsilon}) \cup (\sqrt{1 - \epsilon}, \sqrt{1 + \epsilon}).
$$
So to pick $\lambda$, make $\lambda = 1 - \sqrt{1 - \epsilon}$. This is because $1 - \sqrt{1 - \epsilon} > \sqrt{1 + \epsilon} - 1$.
