Delta-epsilon proof (book example)

[OrbitalPower]

$\lim_{x \to 0}$ x^(1/3)

I know that $\delta = {\epsilon}^3$

the book gives an example:

$\lim_{x \to 2}$ (3x - 2) = 4 and you chose $\delta = \frac{\epsilon}{3}$
so

0 < |x-2| < $\delta = \frac{\epsilon}{3}$

implies

|(3x - 2) - 4| = 3|x-2| < $3 (\frac{\epsilon}{3}) = \epsilon$

so i should get something like:

$| \sqrt[3]{x} - 0 | = | x - 0 | = \epsilon$

But I don't see how you make the connection between $| \sqrt[3]{x} |$ and (|x - 0|) < $\delta$ and I don't see how I can reduce ${\epsilon}^3$ to just epsilon using this style of proof.

 

[Focus]

Think of it this way, start out with$$|x^{\frac{1}{3}}|$$ and try to get this to be $$\leq |x| < \delta$$ (we both know the limit is zero ;)). Your choice of $$\delta$$ seems to be correct. Now given that $$|x| < \delta$$ what is $$|x^{\frac{1}{3}}|$$ less than?

Good luck

 

[Pere Callahan]

You are suspecting that x^1/3 converges to zero as x approaches 0. So how does one show that. First you pick an arbitrary e>0. Then you want to find an d>0 with the following property.
Whenever you take an x which differs from zero by at most d and plug this x into the function x^1/3, you want to get something which differs from the proposed limit, 0, by at most e.

The question is how to choose that d, right?
You want |x^1/3-0 |<d for all |x-0|<d. We can simplify this to
|x^1/3|<e for all |x|<d, right?
Is the first equation not equivalent to |x|<e³? So, how small an x do you have to choose to make sure that this is certified. Well, certainly an x with |x|<e³, right? So what should d be then?

 

[OrbitalPower]

Thanks.

I thought we wanted |x^(1/3) - 0 | < e for all | x - 0| < d. I think I see where to go, though.