What is the logic behind this sequence limit proof?

The question is:

The solution given is:

Is there some mistake with this solution? I understand that the limit is $\frac{3}{2}$, but their reasoning, to me, falls short of proving this.

What I don't understand is why they solved for N and then subbed it back in the equation, immediately after deriving it to find that $\epsilon = \epsilon$. Then they use this trivial fact as the reasoning behind $\frac{3}{2}$ being the limit (as n approaches infinity).

If I would use another number--other than $\frac{3}{2}$ (not the limit as n approaches inifinity)--I could repeat the process and get the same result.

Answer

Let's say we tried to prove the same way that the limit was $1$.

Then

$$|a_n-1| = \frac{n+8}{2n-1}$$

Now we already see a problem - if we try to solve $\frac{n+8}{2n-1}<\epsilon$, what do we get:

$$n+8<\epsilon(2n-1)\\ 8+\epsilon < (2\epsilon-1)n$$

But wait, now for small $\epsilon$, $2\epsilon-1<0$ so this means that:

$$n<\frac{8+\epsilon}{2\epsilon-1}$$

So we get the opposite of what we want, at least for $\epsilon<\frac{1}{2}$. Then, for large $n$, we find that when $n>0>\frac{8+\epsilon}{2\epsilon-1}$, that $|a_n-1|>\epsilon$. So the limit is not one.

It's worth noting that, in the above the middle part of the proof is completely necessary - you could prove this even without that middle part. The middle part is just how you derive the formula for $N$. The actual proof is:

Show

$$\left|a_n-\frac{3}{2}\right|=\frac{17}{4n-2}$$
Given $\epsilon>0$, let $N=\frac{17}{4\epsilon}+\frac{1}{2}$.

Then show if $n>N$ then $\left|a_n-\frac{3}{2}\right|<\epsilon$.

The middle part is just the way you derive $N$, but it isn't necessary for the proof. In other words, of course you've picked exactly the $N$ you needed. That is the point.