Need help proving something.

Show using the definition of a limit that lim (as x approaches c) of (x^3)=(c^3) for all c in R.

Thank you in anticipation.

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- Oct 30th 2006, 04:16 PMFares23Limit Definition
Need help proving something.

Show using the definition of a limit that lim (as x approaches c) of (x^3)=(c^3) for all c in R.

Thank you in anticipation. - Oct 30th 2006, 05:34 PMThePerfectHacker
If $\displaystyle c\geq 0$ then,

For every $\displaystyle \epsilon>0$

There is $\displaystyle \delta>0$

such as,

if $\displaystyle |x-c|<\delta$ is in domain,

then,

$\displaystyle |x^3-c^3|<\epsilon$.

The first inequality can be expressed as,

$\displaystyle c-\delta<x<c+\delta$

And,

$\displaystyle c^3-\epsilon<x^3<c^3+\epsilon$

Thus,

$\displaystyle \sqrt[3]{c^3-\epsilon}<\sqrt[3]{-\epsilon}<x<\sqrt[3]{\epsilon}<\sqrt[3]{c^3+\epsilon}$

If you chose,

$\displaystyle c+\delta=\sqrt[3]{\epsilon}$

$\displaystyle c-\delta=\sqrt[3]{-\epsilon}$

Thus,

$\displaystyle \delta=-c+\sqrt[3]{\epsilon}$

Would make the inequality work can gaurentte the function is defined.

Now you do the second half if, $\displaystyle c<0$.

There got to be a more simple way to do it. But that is the only one that I came up with. - Oct 30th 2006, 09:51 PMFares23
You don't have to write out the whole thing, but what changes for c<0?

- Oct 31st 2006, 05:25 AMThePerfectHacker
After manipulating the epsilon I got,

$\displaystyle \sqrt[3]{c^3-\epsilon}<x<\sqrt[3]{c^3+\epsilon}$

But, since $\displaystyle c> 0$

$\displaystyle \sqrt[3]{\epsilon}<\sqrt[3]{c^3+\epsilon}$

That was what I was trying to do.

But the problem is I just realized that what I did was wrong. Maybe I will try to re-edit what I posted later on. - Oct 31st 2006, 03:35 PMFares23
What did you do wrong in the first part?

- Oct 31st 2006, 05:59 PMThePerfectHacker