1. ## 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.

2. 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
And,
$c^3-\epsilon
Thus,
$\sqrt[3]{c^3-\epsilon}<\sqrt[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.

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

4. 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}
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.

5. What did you do wrong in the first part?

6. Originally Posted by Fares23
What did you do wrong in the first part?
The inequality,
$\sqrt[3]{c^3-\epsilon}<\sqrt[3]{\epsilon}$