Equality of limit and a sequence and its terms Two numbers $a$ and $b$ are equal if and only if for every $\epsilon > 0 $, 
$|a-b| < \epsilon $
A sequence is said to converge to a limit $L$ if for any $\epsilon >0,\ 
 \exists  m \in $ N such that $ |x_n - L| <\epsilon\ for\ all\ n\ge m $
Does the second statement imply that after a certain $m$, all the terms in the sequence are equal to the limit?
 A: No, here is a counterexample. Let $x_n = \frac{1}{n}$. We want to show that it converges to $L = 0$. Let $\epsilon > 0$. Then pick some $m$ such that $m > \frac{1}{\epsilon}$. For all $n \geq m - \frac{1}{\epsilon}$ we have
$$ |x_n - 0 | = | \frac{1}{n} | \leq |\frac{1}{m} | < \epsilon. $$
But, obviously, no $x_n$ is equal to $0$ for any $n$.

The reason why you can't use the definition of equality in the equation $|x_n - L |< \epsilon$ is that the choice of $n$ depends on the choice of $\epsilon$. When taking the limit, you do the following:


*

*Pick an $\epsilon > 0$. Find a lower bound $m_\epsilon$ (corresponding to that value of $\epsilon$.) 

*For all $n \geq m$, verify $|x_n - L| < \epsilon$.


When testing for equality of $x_n$ and $L$, you do the following:


*

*Pick some $x_n$ and $L$.

*Now for all $\epsilon > 0$, verify $|x_n - L| < \epsilon$.


The key here is that $\forall A \exists B$ is not the same as $\exists B s.t. \forall A$. The order of the quantifiers you use here matters. (A similar principle is at play with continuity and uniform continuity.) 
