Proving x^(1/3) is continuous.

1. Oct 24, 2012

luke8ball

1. The problem statement, all variables and given/known data

Prove x^(1/3) is continuous on all of ℝ.

3. The attempt at a solution

I've essentially gotten everything to the following point:

[|x-c|/|x2/3 + (cx)1/3 + c2/3|]<ε

I'm having trouble coming up with a lower bound for the denominator. Any help?

2. Oct 24, 2012

Zondrina

Did it ask you to use the epsilon delta definition? Really?

Why don't you use the definition of the derivative to show that x1/3 is continuous everywhere?

First off, notice that x1/3 is continuous $\forall x \in ℝ$

Now you want to show that its derivative, $\frac{1}{3} x^{- \frac{2}{3}}$ is also continuous $\forall x \in ℝ$.

Notice though, if you pick x = x0 and apply the definition of the derivative to x1/3, you will find that x1/3 has continuous derivatives everywhere except possibly the origin?

Now what happens when you apply the definition of the derivative on x1/3 for x = 0? Does your limit exist?

3. Oct 24, 2012

luke8ball

Thanks for your quick response! Unfortunately, we haven't gotten to the chapter on derivatives yet, and we're only allowed to use an epsilon-delta proof..

4. Oct 24, 2012

Zondrina

So then really what you want to show is :

$\lim_{x→a} x^{\frac{1}{3}} = a^{\frac{1}{3}}$

$\forall ε > 0, \exists δ > 0 \space | \space 0 < |x-a| < δ \Rightarrow |x^{\frac{1}{3}} - a^{\frac{1}{3}}| < ε$

Now start by massaging $|x^{\frac{1}{3}} - a^{\frac{1}{3}}|$ into $|x-a| < δ$

5. Oct 24, 2012

luke8ball

Sorry, I should've showed my earlier steps. I already started working on $|x^{\frac{1}{3}} - c^{\frac{1}{3}}|$.

From there, I multiplied the numerator and denominator to get to [|x-c|/|x2/3 + (cx)1/3 + c2/3|]<ε

Now I have my |x-c|, but I can't come up with a lower bound on the denominator. I've tried a bit of algebraic manipulation, but to no avail.

6. Oct 24, 2012

Zondrina

Ah I see, so. After you multiply by the conjugate you know that |x-c| < δ right? So that mess [|x-c|/|x2/3 + (cx)1/3 + c2/3|] < δ/|x2/3 + (cx)1/3 + c2/3|

( I'm assuming you've done your arithmetic correctly here, don't have time to check at the moment ).

Now you want to ask yourself, how big does :

|x2/3 + (cx)1/3 + c2/3| get?

7. Oct 24, 2012

luke8ball

I'm actually trying to see how small it gets. Hence, I need |x2/3 + (cx)1/3 + c2/3| to be greater than some fixed value.

8. Oct 24, 2012

Klungo

Just so you know. It is ok to choose a two values of x near c to find this upper limit. After all, it is the only place of interest.

9. Oct 24, 2012

luke8ball

Do you mean going ahead and saying δ<1 (or some number) so that |x-c|<1, implying x>(1-c)?

I tried something of that sort but ran into some issues, because the bound still depended on the value of c.

10. Oct 24, 2012

Klungo

Yes.

Let sigma be less than 1. The |x-c|<1.
Thus, c-1<x<1+c. So substitute for which ever x to minimize
|x^(2/3) + (cx)^(1/3) + c^(2/3)|.

Clearly, for c-1 substituted for x is a lower bound of that expression.

Continue from there and also note that I'm assuming the expression is correct.