Limits and convergence - selecting the right epsilon for proofs Show that if limit of sequence $x_n$ goes to $x$ as $n$ goes to $ \infty$ and $ x > a$, then $x_n > a$ all but finitely many $ n$.
I have started it like this:
$x_n > x - \epsilon$ (as  $|x_n- x| < \epsilon$ for all $\epsilon$)
$x_n > a - \epsilon$
From here on, how does one chose the right epsilon to show that $x_n > a$?
 A: It is always best to draw a number line (if possible do it mentally rather than on paper) and position $x, a$ on it: $$------a-----x-----$$ Now things are crystal clear. As $n\to\infty$ the values of $x_n$ will come near $x$ and from the figure above you can see that as one moves near $x$ one gets greater than $a$.
The question is "how near should one go to $x$ in order to become greater than $a$?". Since the distance between $x$ and $a$ is $|x-a|=x-a$ one has to move closer to $x$ by anything less than this amount. Choose any positive $\epsilon$ less than $x-a$ and you are done. In particular you can choose $\epsilon=(x-a) /2$. Note that $\epsilon=x-a$ also works because the inequality used in limit definition is $<$ and not $\leq $. 
A: Let $\epsilon = \frac {x-a}{2}$
Then $x-\epsilon = \frac {x+a}{2} >a$
Thus  $x_n > x-\epsilon \implies x_n >a$
A: If a sequence $x_n$ converges to $x$, by definition it means for every $\epsilon >0$ there exist a $N$ such that $$|x_n-x|<\epsilon\tag{1}$$ for all $n>N$
Now (1) can be rephrased as $x_n\in (x-\epsilon, x+\epsilon)\tag{2}$ for all $n>N$
Take $\epsilon=x-a$ (as $x>a, x-a$ is a positive number and hence the choice of taking $x-a=\epsilon$ makes sense).
By definition there exist a $N$ such that 
$$x_n\in (x-\epsilon, x+\epsilon)$$ for all $n>N$. Now put the value of $\epsilon $ you get $$x_n\in (x-(x-a),x+x-a)=(a,2x-a)$$
Therefore $x_n>a$ for all $n>N$
