Prove that if and if , then there exists a natural number such that for all .

My thoughts

Intuitively, this makes sense. If as then this means that at some point where .

So far I've gotten:

Given then,

Printable View

- February 3rd 2010, 05:35 PMLuxuryAdvanced Calculus - Limit Proof
Prove that if and if , then there exists a natural number such that for all .

**My thoughts**

Intuitively, this makes sense. If as then this means that at some point where .

So far I've gotten:

Given then,

- February 3rd 2010, 05:44 PMDrexel28
- February 3rd 2010, 06:10 PMLuxury
- February 3rd 2010, 06:27 PMDrexel28
- February 22nd 2010, 12:30 AMJeanneLim(x_n)=x and x>0
Since x>0, it is a little easier to simply let x=epsilon.

Substitute in x for epsilon in you inequality.

And since from the definition of the , let K(epsilon) = M.