# Homework Help: Need Help with Formal Definition of Limits

1. Oct 6, 2011

### chez_butt23

1. The problem statement, all variables and given/known data
Limit a$\underline{}n$ as n→∞ = a. Find the limit a, and Determine N so that absolute value(a$\underline{}n$ - a) < $\epsilon$ for all n>N for the given value of $\epsilon$.

The problem that I am working on is:

a$\underline{}n$ = 1/n , $\epsilon$ = 0.01

I'm sure this is very simple, as I am only two weeks into my university's basic calcuus class, but I am not nderstanding what to do. I have also tried going to tutoring and office hours, but my professor only confuses me more with his broken English.

2. Relevant equations

I am not sure what N is. I know that n is the nmber we are currently plugging in. I also know that a$\underline{}n$ is the whatever equation we are using (in this problem it is 1/n), and I know that $\epsilon$ is a margin above and below the limit.

3. The attempt at a solution

I saarted with the equation:
absolute value((a$\underline{}n$) - a) <$\epsilon$

I then plugged in numbers to get:
absolute value ((1/n)-0) < 0.01

After dropping the absolute value (because the limit is zero, and I think I am only solving for positive$\epsilon$), and isolating n, I proceeded to get:
100 < n

I do not know what to do from here. I am not sure what n>N means or how to solve for it. Thank you so much.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 6, 2011

### Petek

N is an unknown quantity that you have to find. Its value depends upon ε. Perhaps thinking of it this way will help: We're playing a game. I give you a specific value for ε. You have to give me back a value for N such that any time that n > N, then 1/n < ε. So, I give you ε = 0.01. You have to find an N such that whenever n > N, then 1/n < 0.01. What value of N would work? You've already done most of the work. You just have to put it together. Hope this helps.

3. Oct 6, 2011

### chez_butt23

Thank you for the reply, I really appreciate it.

Please correct me if I'm wrong, but because n > 100, and because n > N, we could set N = 100. This means that 1/n < 1/N → 1/n < 1/100 → 1/n < ε. Is that seriously the answer, because if so, I want to slap myself in the face right now.

4. Oct 6, 2011

### HallsofIvy

Yes, that is seriously the answer- you may now slap!

Obviously, the limit is 0 so, essentially, you want |(1/n)- 0|< .01. Of course, 1/n- 0= 1/n and since n> 0 that is the same as 1/n< .01 so n> 100. Choose N to be any number greater than or equal to 100 and it follows that if n> 100, then 1/n= |1/n- 0|< 0.01.