Here is the problem:
Using the $ \epsilon$-$N $ definition of limit. Prove that if $ lim \space x_{n} = 1 $, then $ lim \space|5x_{n} - 3| = |5-3| = 2$
We have been given
$$ \tag{1}\forall \space \epsilon >0 \space \exists \space N , \space \forall \space n > N \space |x_{n}-1| < \epsilon $$
We wish to show that some M exists, such that
$$ \tag{2}\forall \space \epsilon > 0 \space \exists \space M, \space \forall \space n > M \space ||5x_{n}-3|-2|<\epsilon $$
Due to the nested inequality, I examine two cases:
When $x \ge \frac{3}{5}; |5x_{n} - 3 | = 5x_{n} - 3$
Using the reverse triangle inequality we get $ |5x_{n} - 3 | - |2| \le |5x_{n} - 5| = 5|x_{n}-1| $. Can I then pick a value? Say $\epsilon ', \epsilon ' > 0, \space where \space \epsilon ' =5\epsilon$, so that I can then say - $ |5x_{n} - 3 | - |2| \le 5 |x_{n} - 1| < \epsilon ' \implies |5x_{n} - 3 | - |2| \le |x_{n} - 1 | < \frac{5\epsilon}{5} = \epsilon$
Thus, giving me the inequality I need to show that $ ||5x_{n} - 3| - 2| < \epsilon $ (I'm taking liberty with the $\le$ here).
I would then use a similar approach to show the second case when $x < \frac{3}{5}$
You are making it too hard on yourself.
If $\epsilon =\frac 25$ then there is a $K$ so that $n > K$ will imply $|x_n - 1| < \frac 25$. That means if $n > K$ then $x_n > \frac 35$.
So just assume $M \ge K$ and we will always have $n> M\implies 5x_n - 3 > 0$ and we can completely ignore worrying about $x_n \le \frac 35$.
......
Now if we want $||5x_n - 3| - 2| < \epsilon$ and we assume $n$ is at least as high as the $K$ above, we want $|5x_n - 5| < \epsilon$.
But $|5x_n - 5| < \epsilon \iff |x_n - 1| < \frac 15 \epsilon$.
Now we know there is an $L$ so that if $n > L$ then $|x_n-1| < \frac 15\epsilon$.
So if we let $M = \max(K,L)$ there $K$ is so that $n > K\implies |x_n-1| < \frac 25$ and $L$ is such that $|x_n -1| < \frac 15 \epsilon$ then if
$n > M\ge K$ then we have $|x_n-1| < \frac 25$ so $x_n > \frac 35$ and $5x_n -3 > 0$ and $|5x_n -3| =5x_n -3$.
And if $n > M \ge L$ then we have $||5x_n -3| - 2| = |5x_n-5| > 5\times \frac 15 \epsilon = \epsilon$.
And that's it. That's all.