Question

prove, using the definition of limit that lim(x approaches -2) x^3=-8

Answer 1

The limit: \[
\lim_{x\to a}f(x) = L
\]is defined as: \[
\forall \epsilon>0\;\exists\delta>0\quad\forall x:\;0\lt |x-a|\lt\delta\implies |f(x)-L|\lt \epsilon
\]That is to say,
For any arbitrary \(\epsilon\) range
You must provide a method of finding a \(\delta\) range such that:
For any \(x\) is in the \(\delta\) range of \(a\):\[
a-\delta \lt x \lt a+\delta
\]Every \(f(x)\) must be within the \(\epsilon\) range of \(L\):\[
L-\epsilon\lt f(x) \lt L+\epsilon
\]In general, the method is some function \(g(\epsilon)\) which returns a valid \(\delta\).

Answer 2

\[
\lim_{x\to (-2)}x^3 = (-8)
\]is defined as: \[
\epsilon>0\;\exists\delta>0\quad\forall x:\;0\lt |x- (-2)|\lt\delta\implies |x^3 - (-8)|\lt \epsilon
\]

Answer 3

\[
\begin{array}{rcl}
\forall \epsilon>0\;\exists\delta>0\quad\forall x:\; \\
\\
0\lt |x- (-2)|\lt\delta &\implies& |x^3 - (-8)|\lt \epsilon \\
0\lt |x + 2|\lt\delta &\implies& |x^3 + 8|\lt \epsilon \\
0\lt |x + 2|\lt\delta &\implies& |x +2||x^2-2x+ 4|\lt \epsilon \\
\end{array}


\]

Answer 4

One thing to take note is that: \[
|x^2-2x+4| = x^2-2x+4

\]Because it is always positive.

Now the trick here is we can't simply say \[
g(\epsilon) = \frac{\epsilon}{x^2-2x+4}
\]Our method \(g\) must be independent of \(x\).

Answer 5

However, one thing we have on our side is that we can let \(\delta\) smaller than it needs to be.

Answer 6

We can also use the minimum function.

Answer 7

Suppose we let \(\delta = 1\):\[
\begin{array}{rcccl}
0 &<& |x+2| &<& 1 \\
-1 &<& x+2 &<& 1 \\
-3 &<& x &<& -1 \\
\end{array}
\\
\]What happens to \(x^2-2x+4\)?\[
(-3)^2-2(-3)+4 = 19 \\
(-1)^2-2(-1)+4 = 5
\]
Remember that we want to minimize \(\delta\)
To do so we must to maximize \(x^2-2x+4\), because it is in the denominator.
When \(1\) is our delta range, we know that \(19\) is the highest \(x^2-2x+4\) will be.

So we let: \[
g(\epsilon) = \frac{\epsilon}{\max(x^2-2x+4)}=\frac{\epsilon}{19}
\]There is one issue here though:
\(19\) is only the max when \(\delta=1\) and thus \(-3<x<-1\).
How about when \(\delta > 1\)?
The solution is the minimum function! Never let \(\delta>1\) be the case!
No one is forcing us to chose \(\delta > 1\). \[
\delta = g(\epsilon) = \min\left(1,\frac{\epsilon}{19}\right)
\]
And this concludes the proof.

Answer 8

@Opcode Tell me what you think.

Answer 9

Which part doesn't make sense or is hard to follow?

Answer 10

I probably should have written: \[
|x+2||x^2-2x+4| < \delta |x^2-2x+4| < \epsilon \implies \delta < \frac{\epsilon}{|x^2-2x+4|}
\]

Answer 11

@abb0t Want to double check?

Answer 12

The last part is how I usually write it. But damn, that's a lot of work! Lol. Nicely done though.

Answer 13

The tricky part of these proofs is you can't let \(\delta\) be a function of \(x\).

Answer 14

If you could, then you would be able to prove false things.

Answer 15

Lol. Let me start from the top.

Answer 16

Where did the \(x^2+2x-4\) come from?

Answer 17

\[
x^3+8 = (x+2)(x^2-2x+4)
\]

Answer 18

\[
(a^3+b^3) = (a+b)(a^2-ab+b^2)
\]

Answer 19

I don't think i'm understanding the min/ max portion of it. Because you're only really concerned with what's going on around x = -2 and you already have something familiar. And I think you already have a term that's familiar on the right. If i'm following your steps correctly.

Answer 20

Well suppose they chose \(\epsilon = 10000\)

Answer 21

We don't want to choose \(\delta = 10000/19\) because it's too big a range.

Answer 22

Essentially it is protecting use from big epsilon having us pick too big a delta.

Answer 23

\[
|x+2| < 10000/19 \implies -10000/19<x < 10000/19-2
\]If we plug \(10000/19\) (which is now allowed) into \(|x^3 + 8| < 10000\)
We find out it is false...

Answer 24

Whoops... \(10000/19-3\) is now allowed, I mean.

Answer 25

That's why we just want to say \(\delta =1\). =]

Answer 26

uh...potato?

Answer 27

The \(\min\) part is really only there for rigor. At some point we need to say "Let's not get any further from \(-2\) than this"

Answer 28

The \(\min\) is just saying let's got get really far away from \(x\) because to doesn't matter past that point.