Question

Prove: lim x^3=8 as x approaches 2, I got the scratch work down but I have no idea how to prove it.

Answer 1

You have to establish that for any \(\epsilon>0\), there's some \(\delta\) such that \(0<|x-2|<\delta\) satisfies \(|x^3-8|<\epsilon\). (If any of this is unfamiliar, refer to the definition of the limit.)

The procedure is to start off with the \(\epsilon\) inequality an derive a proper expression for \(\delta\) in terms of \(\epsilon\). By this, I mean we use the inequality \(|x^3-8|<\epsilon\) to find an expression like this inequality, \(|x-2|<\delta\), where \(\delta\) depends on \(\epsilon\).
\[\begin{align*}
|x^3-8|&=\left|(x-2)(x^2+2x+4)\right|&\text{simply factoring a difference of cubes}\\
&=\color{red}{|x-2|}\left|x^2+2x+4\right|&\text{splitting up the factors}
\end{align*}\]
The red term is what we want to isolate. Okay so far?

Answer 2

so far so good

Answer 3

Alright. Quadratic factors are tricky to work with directly, so to get around them we agree to a bound on \(\delta\). Usually, we would agree to set \(\delta\le1\).

Under this condition, we can find a bound for the quadratic factor, or at least a part of the quadratic factor.

Keep in mind that we're given \(|x-2|<\delta\). Since we set \(\delta\le1\), we get \(|x-2|<1\).

Let's rewrite the inequality so that we can get another inequality in terms of the variable \(x\) (as opposed to \(x-2\)).
\[|x-2|<1~~\iff~~-1<x-2<1~~\iff~~1<x<3\]
So no matter what the value of \(x\) is, it's obvious that its absolute value is smaller than the upper bound of this interval. In other word we can simply write \(|x|<3\).

This is the part that usually confuses people about limit proofs, and I'll admit it's a bit of a hand-wavey explanation. I'm avoiding justifying every step of the process because I'm not sure I can explain it properly...

Anyway, does this make sense to some degree?

Answer 4

Yeah I'm still understanding your explanation up to this part. You're doing great so far!

Answer 5

Great!

Okay, so we've established that \(|x|<3\), but how do we use this information?

Let's go back to our inequality:
\[\begin{align*}
|x^3-8|&=\left|(x-2)(x^2+2x+4)\right|\\
&=|x-2|\color{green}{\left|x^2+2x+4\right|}&\text{focus on the green now...}\\
&\le|x-2|\color{green}{\left(|x^2|+|2x|+|4|\right)}&\text{by the triangle inequality}\\
&=|x-2|\color{green}{\left(|x|^2+2|x|+4\right)}&\text{properties of abs val}
\end{align*}\]
Now we have expressions of \(|x|\) in the inequality which can be replaced using \(|x|<3\):
\[\begin{align*}
|x^3-8|&=\cdots\\
&=|x-2|\left(|x|^2+2|x|+4\right)\\
&<|x-2|\left(3^2+2(3)+4\right)\\
&=19|x-2|...............&<\epsilon
\end{align*}\]
Got it? Be careful with the in/equality symbols. The most recent symbol applies to the general expression. In this case, we have \(|x^3-8|<19|x-2|<\epsilon\).

If we want to isolate \(|x-2|\), we get \(|x-2|<\dfrac{\epsilon}{19}\).

But wait! We can't just say \(\delta=\dfrac{\epsilon}{19}\). We still have to account for the assumption we made earlier, that \(\delta\le1\). So for this \(\epsilon\) expression to work properly, we have to choose the smallest of the two \(\delta\)s.

In other words, we find that \(|x-2|<\delta=\min\left\{1,\dfrac{\epsilon}{19}\right\}\). (\(\min\{a,b\}\) is the smaller of \(a\) and \(b\))

Answer 6

Alright I understand all of this so far, this was as far as I got on my own... but how do you apply the above (scratch work) to do the proof?

Answer 7

You could write the formal proof as follows.
\[\text{For any }\epsilon>0\text{ there exists }\delta=\min\left\{1,\frac{\epsilon}{19}\right\}\text{ such that...}\]
The ... means you'd fill in the blanks with what is essentially the reversed algebra.

As in the definition of the limit, you have to show that \(|x-c|<\delta\) implies \(|f(x)-L|<\epsilon\).

Answer 8

gotcha so I would practically have to repeat most of the work we did earlier?

Answer 9

Yup that's how the proofs tend to look.

Answer 10

Thanks so much! You were an awesome help, I definitely understand the process of these problems better now :D

Answer 11

You're welcome!

Answer 12

nice proof

Answer 13

but too much work