# Prove that (|x-1| < delta) implies (|x^2 - 1| < 1/2)

1. Feb 8, 2015

### Nidhogg

1. The problem statement, all variables and given/known data
"Use the given graph of f(x) = x^2 to find a number delta such that..."

2. Relevant equations
If $([\lim_{x \rightarrow a} f(x)] - L)$ then there exists an $\epsilon$ such that $[0 < |x - a| < \delta] \Rightarrow [|f(x) - L| < \epsilon]$.
Here, f(x) and $\epsilon$ are given. So we have to find a $\delta$ such that $|x-1|<\delta$ implies $|x^{2}-1| < \frac{1}{2}$.

3. The attempt at a solution
I assume that we have to relate $x^{2}-1$ with $\delta$ somehow. I tried to do it like this:
Prove that $(|x-1| < \delta) \Rightarrow (|x^{2}-1|<\frac{1}{2})$.
Choose $\delta$:
$|x^{2}-1|<\frac{1}{2}$
$\Rightarrow -\frac{1}{2}<x^{2}-1<\frac{1}{2}$
$\Rightarrow -\frac{1}{2}+1<x^{2}<\frac{1}{2}+1$
$\Rightarrow \frac{1}{2}<x^{2}<\frac{3}{2}$
$\Rightarrow \sqrt{\frac{1}{2}}<x<\sqrt{\frac{3}{2}}$

But here I get stuck. I feel that I have to relate this to $\delta$, but I don't know how!

As an aside, I can do regular $\delta-\epsilon$ proofs with linear equations now, so by rights this should be easy. But for some reason, it isn't.

2. Feb 8, 2015

### vela

Staff Emeritus
Try factoring $x^2-1$ and find an upper bound for x+1.

3. Feb 8, 2015

### LCKurtz

I don't think the question is asking you to work out a full $\delta,~\epsilon$ argument. Look at your picture and fill in the numbers for the ? marks. Then answer the question. It is there in the picture.

4. Feb 9, 2015

### Nidhogg

D'oh! I didn't even realize that it was that simple. I was overthinking it. Obviously, the two $\delta$ values are $\sqrt{0.5}$ and $\sqrt{1.5}$, which $\approx$ 0.70 and 1.22 respectively. Plug either of these into the original function, subtract one, and take the absolute value, and you're left with .51 and 0.4484 So choose 1.22 for $\delta$ and you're good. I figured this out with a rather long $\delta-\epsilon$ argument below, in which I wandered about and "took the scenic route," as the saying goes (I'm not good at math yet, in case you can't tell), but I did figure it out eventually.

Okay, let's try that. We have $|x^{2}-1|<\frac{1}{2}$. So let's factor: $|(x+1)(x-1)|<\frac{1}{2}$. We can now write $|x-1|<\frac{\frac{1}{2}}{x+1}$, and simplify that to $|x-1|<\frac{1}{2x+2}$. We can now find an upper bound for $|x+1|$ with a quadratic equation. We know that $|x-1| < \frac{1}{2x+2}$, so to find an upper bound we simply set $|x-1| = \frac{1}{2x+2}$, and multiply both sides by $2x+2$ to obtain $|x-1|(2x+2) = 1$. Subtract 1 and simplify the expression to obtain $2x^{2}-2x+2x-2-1 = 0$. Combine like terms to get $2x^{2}-3 = 0$. Now for the quadratic: $(2x^{2}-3 = 0) \Rightarrow (x = \frac{0 \pm \sqrt{0^{2} - (4 \cdot 2 \cdot -3)}}{4})$. This simplifies to $x = \frac{0 \pm \sqrt{24}}{4} = \frac{\pm 2\sqrt{6}}{4} = \frac{\pm \sqrt{6}}{2}$. (notice that $\frac{\sqrt{6}}{2}$ is equal to $sqrt{1.5}$. We now have an upper bound for x. We want a $\delta > 0$, so drop the $\pm$ sign and just use choose any $\delta < \frac{\sqrt{6}}{2}$.
To see this in action, let's choose $x = \frac{\sqrt{6}}{2}$ and use the limit definition, with $<$ changed to $\leq$ because we're finding an upper bound here: $(|x^{2}-1| \leq \frac{1}{2}) \Rightarrow (|\frac{\sqrt{6}}{2}^{2} - 1| \leq \frac{1}{2}) \Rightarrow (\frac{6}{4}-1 \leq \frac{1}{2}) \Rightarrow (\frac{1}{2} \leq \frac{1}{2})$. So take $\delta = 1.22$ and you're good. Success!

5. Feb 9, 2015

### LCKurtz

Those are not the $\delta$ values. They are values of $x$ beyond which you don't want to go.

Remember $\delta$ is the magnitude of the difference between $x$ and $1$ that works.