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 $\displaystyle c\geq 0$ 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.

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

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