Limit Definition

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October 30th 2006, 04:16 PM
Fares23

Limit 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.
October 30th 2006, 05:34 PM
ThePerfectHacker

Quote:

Originally Posted by Fares23

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.

If $c\geq 0$ then,
For every $\epsilon>0$
There is $\delta>0$
such as,
if $|x-c|<\delta$ is in domain,
then,
$|x^3-c^3|<\epsilon$ .

The first inequality can be expressed as,
$c-\delta<x<c+\delta$
And,
$c^3-\epsilon<x^3<c^3+\epsilon$
Thus,
$\sqrt[3]{c^3-\epsilon}<\sqrt[3]{-\epsilon}<x<\sqrt[3]{\epsilon}<\sqrt[3]{c^3+\epsilon}$
If you chose,
$c+\delta=\sqrt[3]{\epsilon}$
$c-\delta=\sqrt[3]{-\epsilon}$
Thus,
$\delta=-c+\sqrt[3]{\epsilon}$
Would make the inequality work can gaurentte the function is defined.

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

There got to be a more simple way to do it. But that is the only one that I came up with.
October 30th 2006, 09:51 PM
Fares23

You don't have to write out the whole thing, but what changes for c<0?
October 31st 2006, 05:25 AM
ThePerfectHacker

Quote:

Originally Posted by Fares23

You don't have to write out the whole thing, but what changes for c<0?

After manipulating the epsilon I got,
$\sqrt[3]{c^3-\epsilon}<x<\sqrt[3]{c^3+\epsilon}$
But, since $c> 0$
$\sqrt[3]{\epsilon}<\sqrt[3]{c^3+\epsilon}$
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.
October 31st 2006, 03:35 PM
Fares23

What did you do wrong in the first part?
October 31st 2006, 05:59 PM
ThePerfectHacker

Quote:

Originally Posted by Fares23

What did you do wrong in the first part?

The inequality,
$\sqrt[3]{c^3-\epsilon}<\sqrt[3]{\epsilon}$