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?

For example $\delta = 1/8 $ solve. in fact,
\begin{eqnarray}
|x^2-1| &=& |x-1||x+1|\\
&\le& |x-1|(|x|+1)\\
&\le & 1/8 \cdot 2 = 1/4 <0,45.
\end{eqnarray}
Because
\begin{equation}
|x| \le |x-1| +|1| \le 2.
\end{equation}
Part 2.
 If $x>1, |x^2-1| = x^2-1$ and $x^2-1<0,45 \Rightarrow x< \sqrt{1,45}$. If $x<1, |x^2-1| = 1-x^2$ and $1-x^2<0,45 \Rightarrow x> \sqrt{0,55} $.Then $\delta = \max (|1-\sqrt{0,55}|,|1-\sqrt{1,45}|)$.
