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

[Nidhogg]

Homework Statement

"Use the given graph of f(x) = x^2 to find a number delta such that..."

Homework 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 $|x^{2}-1| < \frac{1}{2}$.

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.

 

[vela]

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

 

[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.

 

[Nidhogg]

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.

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.

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

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!

 

[LCKurtz]

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.

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

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.
!

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