1. ## Proofs using ε

I'm trying to do some revision for my analysis exam and when it comes to proofs where ε is used I dont understand what these proofs mean.

Like for example
Given $\epsilon > 0, \forall n \in N: \left|{x_n - l}\right| < \epsilon$
which is something to do with the limit of the sequence xn being l

why not use
$\left|{x_n - l}\right| = 0$?

analysis just makes no sense D:

2. Originally Posted by renlok
why not use
$\left|{x_n - l}\right| = 0$?
Because $\left| {\frac{1}{n} - 0} \right| \ne 0,~~\forall n$.

3. In fact, all definitions related to limits contain these $\epsilon$ notations. The idea here is that for all $\epsilon>0$, you can still find values of n larger than some N>0 (this should be added to your statement: Not for all n!)and where the difference between $x_n$ and the limit is less than that $\epsilon$. In other words, the sequence is approaching the limit.
In fact, you can't just say $\left|{x_n - l}\right| = 0$ because you may never find a value of $x_n$ that is equal to l. For example, if $x_n = \frac{1}{n}$, then we claim that the limit of $x_n$ is zero as n goes to infinity. However, you can't give any number n such that $\frac{1}{n} = 0!$. In fact, for any given $\epsilon$, look at $|\frac{1}{n}-0|=\frac{1}{n}$ which can be made less than $\epsilon$ if you choose sufficiently large n, namely $n>N = \frac{1}{\epsilon}$. This proves that the limit is zero! That is, for any given $\epsilon>0$, you can find a natural N such that if $n>N$, then $|\frac{1}{n}-0|<\epsilon$
hope this helps!