Uniqueness of a Limit epsilon divided by 2? I have been reading about this theorem in a book called 'Calculus: Basic Concepts for High-schools', it is a very good book (so far) and I can highly recommend it.
Well the author goes on to prove that: 'A convergent sequence has only one limit', which is quite intuitive in  geometrical sense..
However in the beginning of his proof (which is a proof by contradiction) he goes on to state:
"Consider a convergent sequence with two limits a1 and a2 and select a value for epsilon < (absolute value of)a1-a2/2.
First of all why does epsilon have to be smaller than the value on the right of the inequality?
Where does the author derive this inequality from? Is it just some arbitray inequality?
Why does the distance between a1 and a2 have to be divided by 2?
Finally any other explanations/proofs for a more intuitive understanding of the uniqueness of a limit are welcome.
(Excuse me, for not using LATEX it is a thing I'm not yet familiar with..)
 A: If the sequence $(a_n)_{n \in \Bbb{N}} \in \Bbb{R}$ is convergent in $\Bbb{R}$, then there exists a number $a \in \Bbb{R}$ and for every $\varepsilon  > 0$ there exists a number $N$ such that $\left| a_n - a\right | < \varepsilon$ whenever $n > N$.
So suppose the sequence has two limits, $a_1$ and $a_2$. Since the sequence converges to $a_1$, we can find an $m > N$ such that $\left| a_m - a_1\right | < \frac{\varepsilon}{2}$. Remember, if the sequence converges, we can find a number so that the distance between the limit and the terms of the sequence is less than any given positive number, including $\frac{\varepsilon}{2}$.
Since we assumed the sequence also converged to $a_2$, we can do the same thing with that: we can find a $r > N$ such that $\left| a_r - a_2\right | < \frac{\varepsilon}{2}$.
Now, let $M = max (r,m)$. Then, for any $k > M$, both sequences will be within $\frac{\varepsilon}{2}$ of their limits. 
So, now we can show that the limits are both arbitrarily close to one another:
$\left| a_1 - a_2\right | = \left| a_1 - a_k + a_k- a_2\right | \leq \left| a_1 - a_k \right |+ \left|a_k- a_2\right | < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$.
The only trick we used was adding zero ($0 = -a_k + a_k$) and the triangle inequality ($|a + b| \leq |a| + |b|$) for all real numbers $a$ and $b$. So, the conclusion is that the distance between the two limits is smaller than any given arbitrary positive number. The only way this is possible is if they are the same number. So, that means the sequence has one limit which is therefore a unique limit.
A: If a sequence $x_n$ in $\mathbb R$ converges to $a_1$ and $a_2$ with $a_1 \neq a_2$, then there would exist an $N$ with 
$$
 \lvert x_N - a_i \rvert < \epsilon
$$ for $i=1,2$ and $\epsilon = \frac{\lvert a_1 - a_2 \rvert} 2$. This cannot be true since the invervals $(a_i-\epsilon,a_i+\epsilon)$ are disjoint.
A: Here is a rough and hopefully illuminating explanation: if a sequence $(u_n)$ converges to $a_1$, this means any neighbourhood $(a_1-\varepsilon, a_1+\varepsilon)$ of $a_1$ contains all terms $u_n$ of the sequence if $n$ is large enough, say $n\ge N_1$ ($N_1$ depends on $\varepsilon$, of course).
Similarly, if it converges to $a_2$, any neighbourhood of $a_2$ contains all $u_n$s if $n$ is large enough, say $n\ge N_2$ (not necessarily the same as $N1$).
Now, if $n $ is larger than $N_1$  and than $N_2$, all $u_n$s will be in both neigbourhoods $(a_1-\varepsilon, a_1+\varepsilon)$ and  $(a_2-\varepsilon, a_2+\varepsilon)$, which is plainly impossible if $\varepsilon$ was chosen so small  that the two neighbourhoods are disjoint — for instance if we choose $\;\varepsilon<\dfrac{\lvert a_1-a_2\rvert}2$
A: The problem you face is because of $\epsilon$ symbol. Suppose the two limits are $1$ and $2$. This means $a_{n}$ is near $1$ and also near $2$ for large $n$. If you get too close to $1$ then it is obvious that you are not close to $2$. The argument holds even if $1, 2$ are replaced by general numbers $a, b$ provided $a \neq b$.
But to demonstrate this fact (or to prove this fact) you need to quantify how close to one limit $a$ you want $a_{n}$ to be in order to get a bit far from $b$. The best approach is to consider a number which is equidistant from $a, b$. This is the mid point $c = (a + b)/2$. Consider them on real line (let $a$ be the smaller number):
$----a--------c--------b-----$
Clearly if you want your $a_{n}$ to be near $a$ and far from $b$ the only way is to keep the $a_{n}$ left of $c$ like the following
$----a----a_{n}----c--------b-----$
So the distance between $a_{n}$ and $a$ should be less than the distance between $a$ and $c$ i.e. $c - a = (b - a)/2$. This distance is what we quantify by $\epsilon$. Hence we need $\epsilon < (b - a)/2$.
Thinking in terms of real number line is such a great help to visualize the basic concepts related to inequalities and while learning calculus it is better to focus on this aspect rather than the boring $\epsilon, \delta$ stuff.
