Given $\epsilon$>0, and we choose $\delta$ =min of $\left\{1,\frac{\epsilon}{9}\right\}$ Then is it true that if 0 < |x-1|< $\delta$, then |x$^2$-1|< $\epsilon$.>
Now |x$^2$-1| <$\epsilon$ is |x+1||x-1| < $\epsilon$
And I know that if 0 < |x-1|< $\delta$ then 0 < |x+1|< $\delta$+2. is this step right?
Now this means either 0<|x+1|< 1+2, or 0<|x+1| < $\frac{\epsilon}{9}$+2
I'm kind of stuck now because what I want to do next will get quite messy.
If $\epsilon >9$ then $\delta=\min\{1,\frac{\epsilon}{9}\}=1$. Thus $0<|x-1|<1$, and we see that $x\in (0,2)$ (why?)
Consider $|x^2-1|$. For $x\in (0,2)$, $|x^2-1|\in (-1,3)$ (why?)
But we supposed $\epsilon >9$, thus we have shown $|x^2-1|<\epsilon$.
Now suppose instead that $\epsilon\leq 9$, then $\delta =\min\{1,\frac{\epsilon}{9}\}=\frac{\epsilon}{9}$.
Thus we have $|x-1|<\frac{\epsilon}{9}$. Note that $\frac{\epsilon}{9}\leq 1$. (why?)
Observe that for $x=1$, $|x-1|=0$, and since $x-1$ is a line with slope $1$, it follows that:
$|x-1|<\frac{\epsilon}{9}$ imples $x\in (1-\frac{\epsilon}{9},1+\frac{\epsilon}{9})$ (why?)
Now $x^2-1$ is a quadratic, hence graphically it is a parabola. It is centered at $0$ with vertex $(0,-1)$ and concave up ('opens up'). It also has the solution $(1,0)$. Therefore $|x^2-1|$ graphically looks like the parabola $x^2-1$ with the part below the $x$-axis flipped above the $x$-axis instead.
Note that the interval $(1-\frac{\epsilon}{9},1+\frac{\epsilon}{9})\subset (0,2)$ (why?)
We intend to show that, for any $x\in (1-\frac{\epsilon}{9},1+\frac{\epsilon}{9})$, $|x^2-1|<\epsilon$ (given that $\epsilon\leq 9$)
Since $0<1-\frac{\epsilon}{9}<1$, we know from the previous discussion of the graph of $x^2-1$ that $(1-\frac{\epsilon}{9})^2-1\leq 0. $ Thus $|(1-\frac{\epsilon}{9})^2-1|=-(1-\frac{\epsilon}{9})^2+1=1-(1-\frac{2\epsilon}{9}+\frac{\epsilon^2}{81})=\frac{2\epsilon}{9}-\frac{\epsilon^2}{81}$.
Is $\frac{2\epsilon}{9}-\frac{\epsilon^2}{81}<\epsilon$? (given that $\epsilon\leq 9$)
Yes because $\frac{2\epsilon}{9}-\frac{\epsilon^2}{81}<\epsilon \Leftrightarrow -\frac{7\epsilon}{9}-\frac{\epsilon^2}{81}<0$ and the expression on the right is true because for $\epsilon\in (0,9)$, $-\frac{7\epsilon}{9}<0$ and $-\frac{\epsilon^2}{81}<0$. Therefore we have shown that $\frac{2\epsilon}{9}-\frac{\epsilon^2}{81}<\epsilon$ as desired.
Now we need to deal with the case of $1<1+\frac{\epsilon}{9}<2$. Recall the discussion about the graph of $|x^2-1|$. We know that $(1+\frac{\epsilon}{9})^2-1\geq0$, hence $|(1+\frac{\epsilon}{9})^2-1|=(1+\frac{\epsilon}{9})^2-1=\frac{2\epsilon}{9}+\frac{\epsilon^2}{81}$.
Now we ask, is $\frac{2\epsilon}{9}+\frac{\epsilon^2}{81}<\epsilon$? This is true if and only if $-\frac{7\epsilon}{9}+\frac{\epsilon^2}{81}<0$, which is true if and only if $\frac{\epsilon^2}{81}<\frac{7\epsilon}{9}$. Observe that for $\epsilon\in(0,9)$, $\frac{\epsilon^2}{81}$ is a parabola in $\epsilon$ that goes through $(0,0)$ and $(9,1)$ and is concave up, hence it is bounded above by the line that goes through those points, which has slope $\frac{1}{9}$. This is less than the $\frac{7}{9}$ slope of $\frac{7\epsilon}{9}$ which also has goes through the point $(0,0)$, hence $\frac{\epsilon^2}{81}<\frac{\epsilon}{9}<\frac{7\epsilon}{9}$. Hence we have shown $\frac{2\epsilon}{9}+\frac{\epsilon^2}{81}<\epsilon$ as desired.
Finally, to complete our argument for any $x\in (1-\frac{\epsilon}{9},1+\frac{\epsilon}{9})$, we note that, by the previous discussion of the graph of $|x^2-1|$, $|x^2-1|$ is monotone decreasing on $(1-\frac{\epsilon}{9},0]$, and monotone increasing on $[0,1+\frac{\epsilon}{9})$. Therefore, for any $x\in (1-\frac{\epsilon}{9},0]$, $|x^2-1|<\frac{2\epsilon}{9}-\frac{\epsilon^2}{81}<\epsilon$, thus $|x^2-1|<\epsilon$. Similarly for $x\in [0,1+\frac{\epsilon}{9})$, we have $|x^2-1|<\frac{2\epsilon}{9}+\frac{\epsilon^2}{81}<\epsilon$, thus $|x^2-1|<\epsilon$ as required, and our proof is finally complete (more tedious and complicated that I had hoped, but it works).
