Basic analysis - sequence convergence I'm taking a course entitled "Concepts in Real Analysis," and I'm feeling pretty dumb at the moment, because this is obviously quite elementary...
The example in question shows $\lim_{n\to\infty} \frac{3n+1}{2n+5}=\frac{3}{2}$, and, setting $\left|\frac{3n+1}{2n+5}-\frac32\right|=  \frac{13}{4n+10}$, choosing $N>\frac{13-10\varepsilon}{4\varepsilon}$ and $n\ge N$.
Fine.
My question is this: I don't understand why this isn't circular reasoning. I can subtract anything whatsoever from $\frac{3n+1}{2n+5}$, and with a little algebra I can have a statement $n> f(\varepsilon)$, even if I already know the limit and deliberately choose a value for $N$ which disagrees with it, and then I could claim that any $n>f(\varepsilon)$ whatsoever satisfies the criteria for convergence.
I'm sorry I couldn't make the math prettier, but I'm going crazy here. Can anyone help?

It may be a little subtle, but it doesn't actually work unless you pick the limit.
To see this, lets write our limiting value as
$$ \left|\frac{3n+1}{2n+5} - \left(\frac{3}{2} + \delta\right)\right| < \epsilon $$
for some $\delta\neq 0$. Simpilfying, we have
$$ \left| \frac{-13 - (2n+5)\cdot 2\delta}{4n+10} \right| < \epsilon. $$
If
$$ \delta<-\frac{13}{2(2n+5)}<0, $$
the number in the absolute value is positive and we have
$$ \frac{-13 - (2n+5)\cdot 2\delta}{4n+10} < \epsilon. $$
Rewrite this as
$$ n > \frac{10\epsilon - 13 - 10\delta}{4\epsilon + 4\delta}. $$
What happens if we choose $\epsilon = -\delta > 0$? What can we choose for $N$ in the definition of convergence?
On the other hand, if
$$ \delta \geq -\frac{13}{2(2n+5)}, $$
then we have
$$ \frac{13 + (2n+5)\cdot2\delta}{4n+10} < \epsilon, $$
which can be written as
$$ n > \frac{13 - 10\epsilon + 10\delta}{4\epsilon - 4\delta}. $$
What happens for $\epsilon = \delta$?
We have to be a little careful with this last case though. If
$$ - \frac{13}{2(2n+5)} \leq \delta < 0, $$
then this choice of $\epsilon$ is negative, so doesn't contradict the definition. However, in this case we can fall back on the "for all" $n\geq N$ part of the definition of convergence. Note that
$$ \frac{13}{2(2n+5)} $$
gets smaller and smaller as $n$ gets larger, so no such choice of $\delta$ would work for all $n\geq N$, so these won't be possible chioces of $\delta$ anyway!
To try and have a happy ending, note that when you pick the limit (i.e., $\delta = 0$) then as you note above, we have the restriction
$$ n > \frac{13 - 10\epsilon}{4\epsilon}, $$
and the problem that arises for other choices can't actually happen here!
