Proving x^(1/3) is continuous.

[luke8ball]

Homework Statement

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

The Attempt at a Solution

I've essentially gotten everything to the following point:

[|x-c|/|x^2/3 + (cx)^1/3 + c^2/3|]<ε

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

Thanks in advance!

 

[STEMucator]

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

Why don't you use the definition of the derivative to show that x^1/3 is continuous everywhere?

First off, notice that x^1/3 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 x^1/3, you will find that x^1/3 has continuous derivatives everywhere except possibly the origin?

Now what happens when you apply the definition of the derivative on x^1/3 for x = 0? Does your limit exist?

 

[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..

 

[STEMucator]

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..

So then really what you want to show is :

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

So apply your definition :

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

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

 

[luke8ball]

So then really what you want to show is :

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

So apply your definition :

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

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

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|/|x^2/3 + (cx)^1/3 + c^2/3|]<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.

 

[STEMucator]

Ah I see, so. After you multiply by the conjugate you know that |x-c| < δ right? So that mess [|x-c|/|x^2/3 + (cx)^1/3 + c^2/3|] < δ/|x^2/3 + (cx)^1/3 + c^2/3|

( 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 :

|x^2/3 + (cx)^1/3 + c^2/3| get?

Apply your triangle equality now.

 

[luke8ball]

I'm actually trying to see how small it gets. Hence, I need |x^2/3 + (cx)^1/3 + c^2/3| to be greater than some fixed value.

 

[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.

 

[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.

 

[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.