A confusing definition of limit of a sequence

jens.w
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Homework Statement



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.

Homework Equations





The Attempt at a Solution



Well the whole thing is just a big mess in my head, so the questions I am 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?
 
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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.
 
|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.
 
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!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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