<s>[INST] Understanding how to use $\epsilon-\delta$ definition of a limit I finally understand the geometric intuition behind the $\epsilon-\delta$ definition of a limit, which is actually quite neat:

But I'm having trouble actually using the definition to come to a conclusion.
For (a solved) example, to prove that $\lim_{n\rightarrow \infty} \frac{3n+5}{2n+7} = \frac{3}{2}$, the following solution is given:
Proof:
Let $a_n = \frac{3n+5}{2n+7}$. Then, $\left | a_n-\frac{3}{2} \right |=\frac{11}{2(2n+7)}<\frac{3}{n}$. Given $\epsilon > 0$, choose $n_0 \in \mathbb{N}$ such that $n_0 > \frac{3}{\epsilon}$. Then, for all $n \ge n_0, \left | a_n-\frac{3}{2} \right | < \epsilon $. Therefore, $\lim a_n = \frac{3}{2}$.
I'm not quite sure I understand this proof. Namely: 


*

*Where did $\frac{3}{n}$ come from?

*I guess we just replaced n with $\epsilon$ in the third sentence.

*How did the fourth sentence follow from the third, and why does that show that 3/2 is the limit?



So I tried it myself on another problem, where $a_n = \frac{2n+5}{6n-3}$, and the limit being $\frac{1}{3}$:
Let $a_n = \frac{2n+5}{6n-3}$. Then, $\left | a_n-\frac{1}{3} \right |=\left | \frac{6n-6n+15-3}{3(6n-3)} \right |= \left | \frac{12}{3(6n-3)} \right |= 4\left | \frac{1}{6n-3} \right | < \epsilon$. Dividing by 4, $\left | \frac{1}{6n-3} \right | < \frac{\epsilon}{4}$.
And that's as far as I logically get... I assume the next step is to choose an $n_0 \in \mathbb{N}$ such that $n_0 > \frac{\epsilon}{4}$, and isolating epsilon, $\epsilon < 4n_0$.
but as you can tell, I'm really lost.
 [/INST] HINT:
Note that
$$\left|a_n-\frac{1}{3}\right|=\left|\frac{2n+5}{6n-3}-\frac{1}{3}\cdot\frac{2n-1}{2n-1}\right|=\left|\frac{2n+5-(2n-1)}{6n-3}\right|=\left|\frac{6}{6n-3}\right|=\left|\frac{2}{2n-1}\right|$$
Taking $n_0\ge\max\left(\frac{1}{\epsilon}+1,2\right)$ we have the inequality, since
$$n\ge n_0\quad\Longrightarrow\quad\left|a_n-\frac{1}{3}\right|=\left|\frac{2}{2n-1}\right|<\left|\frac{2}{2n-2}\right|=\frac{1}{n-1}<\frac{1}{1/\epsilon}=\epsilon$$
</s>