# Homework Help: Help! Espilon-Delta limits

1. Sep 24, 2010

### Pete.Co.Lust

1. The problem statement, all variables and given/known data
I have already read a tons of definitions about Epsilon-Delta limit proof. But i am still stucking in some places...
E.g.) Prove it by using Espilon-Delta method, lim (x->4) x^3=64

2. Relevant equations

|x-4|<delta delta>0
|x^3|<espilon espilon>0

3. The attempt at a solution

I am kinda know what I should do, I changed |x^3-64| into |(x-4)|(x^2+4x+16)| and my objective afterward is to ensure| (x^2+4x+16)| is "small".
But i just dunno what I should do afterwards... Should I just Sub x=4 and then find out |(x^2+4x+16)|<48?

Plx help me :( ThX!! :)

2. Sep 24, 2010

The point is not in showing that | (x^2+4x+16)| is "small, but that it is "bounded" when x is close to 4. So, restrict yourself to the region, for instance, 3<x<5. What can be said about | (x^2+4x+16)| in this region? Is it always smaller than some fixed number?

3. Sep 24, 2010

4. Sep 24, 2010

### Pete.Co.Lust

Alright I got it. So u mean is to find out the max. number when x is "close" to 4? Afterwards we can state that |x-4|<1 (Because x is "close" to 4) and we can find the max. number of x is 5. we subsitute 5 into the equation and we have (5)^2+(4)(5)+16=61. And it would be espilon/61 afterwards?

If its espilon/61, we write this "delta=min{1,e/61}"? What does "delta=min{1,e/61}" actually means? ThankS

The link is extremely useful for me to solve this problem. I found out the "solution"
(Which i dunno its right or not...) right there. Many thanks.

5. Sep 24, 2010

min{1,e/61} means eiher 1 or e/61, whichever happens to be smaller.

6. Sep 24, 2010

### Pete.Co.Lust

So what do u mean is that if i pick e=62, the output will be 1 and if i pick e=60, the output will be 60/61? min{1,e/61}

7. Sep 24, 2010