Not quite sure if the title wording is correct.
Only using the definition of convergence, prove that if $(x_n) \rightarrow 2$, then $\left( \frac{2x_n - 1}{3} \right) \rightarrow 1$. (Understanding Analysis, ABbott 2nd ed. 2.3.3)
I know that if a sequence converges, then for any choice $\epsilon > 0$, we can find an $n \geq N$ such that $\|a_n - a\| < \epsilon$. Since we know that $(x_n) \rightarrow 2$, that implies the existence of $\| x_n - 2\| < \epsilon$. I'm not entirely sure where to go from here. I manipulated the sequence it wanted me to show the proof for, so that I have
$$\|a_n - a\|= \|\frac{2 x_n - 1}{3} - 1\| = \|\frac{2 x_n - 1 - 3}{3}\| = \|\frac{2}{3}x_n - \frac{4}{3}\| = \frac{2}{3} \|x_n - 2\|$$
From this, I think we say that there exists an $N \in \mathbb{N}$ such that for all $n \geq N$, we have $\|x_n - 2\| < \frac{3}{2} \epsilon$?
I'm not even sure if my though process correct, but if so, I'm still completely stuck.
In this problem, I am assuming that the following analysis is off-limits:
$$\lim_{n \to \infty} x_n = 2 \iff \lim_{n \to \infty} 2x_n = 4 \\ \iff \lim_{n \to \infty} (2x_n - 1) = 3 \iff \lim_{n \to \infty} \frac{2x_n - 1}{3} = 1.$$
So, attacking from scratch, I surmise that the problem composer intended the following:
Given :
$~\forall ~\epsilon_1 > 0, ~\exists ~N \in \Bbb{Z^+}~$ such that for all $~\displaystyle n \geq N, ~\left| x_n - 2 ~\right| < \epsilon_1.$
To prove :
$~\forall ~\epsilon_2 > 0, ~\exists ~N \in \Bbb{Z^+}~$ such that for all $~\displaystyle n \geq N, ~\left| \frac{2x_n - 1}{3} - 1 ~\right| < \epsilon_2.$
My standard approach to this type of problem is:
Given any function $~f(x),~$ and any positive real number $~r,~$
the assertion $~|f(x)| < r~$ will be true if and only if $-r < f(x) < r.$
Work backwards. That is, analyze the assertion that you are trying to prove, and see what relationship is required between $~\epsilon_2~$ and $~N.$ Then, try to use the given premise to establish that relationship.
Given any positive value $~\epsilon_2,~$ you are being asked to prove that there exists a positive integer $~N,~$ such that for any $~n \geq N$:
$$\left| \frac{2x_n - 1}{3} - 1 ~\right| < \epsilon_2 \tag1 $$
$$\iff -\epsilon_2 < \frac{2x_n - 1}{3} - 1 < \epsilon_2$$
$$\iff 1 - \epsilon_2 < \frac{2x_n - 1}{3} < 1 + \epsilon_2$$
$$\iff 3 \times (1 - \epsilon_2) < 2x_n - 1 < 3 \times (1 + \epsilon_2)$$
$$\iff 3 - 3\epsilon_2< 2x_n - 1 < 3 + 3\epsilon_2$$
$$\iff 4 - 3\epsilon_2 < 2x_n < 4 + 3\epsilon_2$$
$$\iff \frac{4 - 3\epsilon_2}{2} < x_n < \frac{4 + 3\epsilon_2}{2}$$
$$\iff 2 - \frac{3}{2} ~\epsilon_2 < x_n < 2 + \frac{3}{2} ~\epsilon_2. \tag2 $$
This means that you are required to prove that $ ~\forall ~\epsilon_2 > 0, ~\exists ~N \in \Bbb{Z^+} ~$
such that for all $~n \geq N,~$ the assertion in (2) holds.
Note that this interpretation of the problem is permitted only because, at each step of the way, going from the assertion in (1) to the assertion in (2), you were able to specify the $~\iff~$ symbol.
That is, the assertion at each step holds if and only if the assertion in the previous step holds. Therefore, if you can prove that the assertion in (2) holds, you are done.
So, the problem reduces to determining how to use the given premise to establish the assertion in (2) above. This may be done as follows:
From the given premise, you know that
$~\forall ~\epsilon_1 > 0, ~\exists ~N \in \Bbb{Z^+}~$ such that for all $~\displaystyle n \geq N:$
$$\left| x_n - 2 ~\right| < \epsilon_1. \tag3 $$
$$\iff -\epsilon_1 < x_n - 2 < \epsilon_2 $$
$$\iff 2 - \epsilon_1 < x_n < 2 + \epsilon_1. \tag4 $$
Just as before, you are able to specify the $~\iff~$ relationship, when going from the assertion in (3) to the assertion in (4). So, the assertion in (4) represents what you can prove, and the assertion in (2) represents what you need to prove.
At this point, in my opinion, the only way to jump from the assertion in (4) to the assertion in (2) is to realize that for any $~\epsilon_2 > 0,~$ the premise allows you to specify that $~\epsilon_1 = \dfrac{3}{2} ~\epsilon_2.$
This is permitted, because inherent in the given premise is the understanding that it holds true for any value $~\epsilon_1 > 0.~$
So, once you specify that the given premise must hold for the particular value $~\epsilon_1 = \dfrac{3}{2} ~\epsilon_2,~$ then you have successfully made the assertion in (4), which is proven, exactly equivalent to the assertion in (2), which needs to be proven.
This means that you are now done.
Addendum
For what it's worth, I agree with the comments following the original posting that indicated that the original posting is fine, as is. This begs the question, why go through all of the tedious steps. There are two reasons for this:
For one thing, similar to my assumption that the analysis at the very start of my answer is off limits, it is unclear whether the analysis within the original posting, which manipulated (absolute value) $~~~| ~\text{expressions} ~| ~~~$ is permitted.
As a second point, absolute value expressions can be tricky. For more complicated problems, when you want to be sure that each step that you are taking is valid, the (tedious) approach that I took may be helpful.
As a side issue, I suspect that the problem composer may be looking for some explicit discussion between (for example) $~\epsilon_1~$ and $~\epsilon_2~$ that justifies the problem solver (in effect) jumping from the assertion in (4) to the assertion in (2).