# Homework Help: A confusing definition of limit of a sequence

1. Feb 2, 2012

### jens.w

1. The problem statement, all variables and given/known data

I'm having an enormously hard time wrapping my head around the following definition, which is using some concepts that keep showing up in other definitions and theorems.

I'll state the definition and then i'll ask about the parts that i don't understand:

We say that $$limx_{n}=L$$ if for every positive number $\epsilon$ there exists a positive number N = N($\epsilon$) such that $$\left | x_{n}-L \right |< \varepsilon$$ holds whenever n $\geq$ N.

2. Relevant equations

3. The attempt at a solution

Well the whole thing is just a big mess in my head, so the questions im able to formulate are:
What is this epsilon number? How is it related to N? Why is $$\left | x_{n}-L \right |< \varepsilon$$ when n $\geq$ N?

If these are irrelevant questions, can you please try to explain this definition from another viewpoint, maybe even graphically?

2. Feb 2, 2012

### Nytik

This isn't quite the definition I learned for limits, but it is very similar. I eventually wrapped my head around it, I'll try my best to explain here...

The number L is what the sequence approximates to as 'n' gets larger and larger.
Use the sum of the sequence (1/2)+(1/4)+(1/8)+... as an example. We know the limit of this as n→∞ is 1.
Take an arbitrarily small epsilon, e.g. 0.1 or something. Can we find an N such that for all n≥N, the difference between our approximation and L=1 is less than 0.1? So Xn would have to be 0.9 or more; in this case N is 4 as 1/2+1/4+1/8+1/16 = 0.9375, and 1-0.9375 is less than our arbitrary epsilon.
It is easy to see that, no matter how small we choose epsilon to be, we will be able to find an N such that the approximation is close enough to L. This is the definition of a limit.

I hope that helped.

3. Feb 2, 2012

### HallsofIvy

$|x_n- L|$ measures how close $x_n$ is to L. Saying there exist N such that if n>N, $|x_n- L|< \epsilon$ means "we can make $x_n$ arbitrarily close to L (less than distance $\epsilon$ from L) by going far enough in the sequence.

4. Feb 6, 2012

### jens.w

Yes, yes, yes. Nytik you made it much clearer. Especially with the example.
Also thank you HallsofIvy.

Now it actually makes sense to put those demands on the sequence, to be comfortable with saying that it will approach a limit. Thank you!