Why do we pick $n_1$ and $n_2$ instead of just $n$? I'm reading Ghorpade/Limaye's: Course in Calculus and Real Analysis: There is this proof that the limit is unique:



I am confused with a small detail: Why do we assume that there is $n_1$ and $n_2$ instead of just $n$? I mean, we are talking about the same sequence $(a_n)$, taking $n_1,n_2$ seems to mean that we are talking about the hypothetical sequences in which $n_1=6$ and $n_2=4$:
$$a_1\quad a_2\quad a_3\quad a_4\quad a_5\quad a_6$$
$$a_1\quad a_2\quad a_3\quad a_4$$
But it doesn't seems to make sense because a sequence is defined as $a_n : \Bbb{N}\to \Bbb{R}$, shouldn't we expect that $n_1=n_2$? Or this process is made to give the idea that we are taking the best bound possible?
 A: Short answer: The value of $n_1$ obtained from the definition of $a_n \to a$ depends on $a$, and the value of $n_2$ obtained from the definition of $a_n \to b$ depends on $b$, so the two values $n_1$ and $n_2$ can't be assumed to be equal.

Long answer: For a fixed sequence $(x_n)$ and a fixed real number $x$, the definition of the statement $x_n \to x$ says that for all $\varepsilon > 0$ there exists $N$ such that a particular condition holds, namely that $|x_n - x| < \varepsilon$ for all $n \ge N$.
Here you have a fixed sequence $(a_n)$ and fixed real numbers $a,b$. Moreover, you've fixed a particular value of $\varepsilon > 0$, namely $|a-b|$.


*

*The definition of $a_n \to a$ tells you that there exists $N$ such that some condition holds, namely that $|a_n-a| < \varepsilon$ for all $n \ge N$; and

*The definition of $a_n \to b$ tells you that there exists $N$ such that another condition holds, namely that $|a_n-b| < \varepsilon$ for all $n \ge N$.


The issue is that these definitions are independent of each other, since $a$ and $b$ are (by assumption) not equal. This means that the two values of $N$ cannot be taken to be equal. Indeed, $N$ can be thought of as a threshold for being close to the limit: the threshold for being close to $a$ might be distinct from the threshold for being close to $b$.
Thus, the value of $N$ given by the first definition is called $n_1$ in your proof; and the value of $N$ given by the second definition is called $n_2$.
These distinct values are then reconciled by taking their maximum, since if you're beyond both the threshold for being close to $a$ and beyond the threshold for being close to $b$, then you must be beyond both thresholds.
A: This simply follows from the definition of limit.
Saying that $a_n\to a$ means that for every $\varepsilon>0$ there exists $N\in \mathbb N$ such that $|a-a_n|<\varepsilon$ for every $n>N$.
If we suppose in all generality that $a_n\to a$, $a_n\to b$ we have no reason to suppose that the "first" $N$ is equal to the "second" $N$; we are merely supposing that two such natural numbers exist. 
From there, it is easy to show that the $\varepsilon$ property holds for the maximum of these numbers, which is exactly what you are saying: the basic statement is somewhat weaker.
A: If the sequence could converge to different values, $a$ and $b$ then it wouldn't necessarily be true that $n_1 = n_2$. There are good answers here that explain why. 
But, to clarify what $n_1$ and $n_2$ are, they are the point(s) in the sequence beyond which $|a_n - a| \lt \frac{\epsilon}{2}$ and $|a_n - b| \lt \frac{\epsilon}{2}$ respectively. So, for $n_1$, say the  relevant terms in the sequence are 
$
a_{n_1}, a_{n_1 + 1}, a_{n_1 + 2}, \ldots
$
All of which are within $\frac{\epsilon}{2}$ of the limit, $a$. The terms we are talking about are not 
$
a_1,a_2,\ldots, a_{n_1}
$
i.e. the the first $n_1$ (or $n_2$) terms of the sequence. 
(The way I read your question it seemed like there may be some confusion on this point. Sorry if that's not what you were saying.)
A: I suppose it is something more fundamental that confuses you. In mathematics,  saying "let $a,b$ be real numbers" does not presume $a\neq b$! Using different symbols just to allow of every possibility. So $a$ may or may not be $= b$. You can think of the statement as "take two real numbers $a,b$ out of $\mathbb{R}$ one at a time with replacement"; so here certainly $a$ may or may not be equal to $b$, and $a,b$ simply represent the first real number and the second real number respectively.
I guess you can solve your problem now. It is just that the author wants to allow of the possibility of $n_{1}=n_{2}$ as well as that of $n_{1}\neq n_{2}$, which follows from the possibility that $a=b$ or that $a\neq b$. 
