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.

Part (1) Essentially whoever wrote the proof wants to get an $n$ in the denominator (to be explained shortly). Note that $\frac{11}{2(2n+7)}=\frac{11}{4n+14}<\frac{11}{4n}<\frac{3}{n}$, so that we got a nice bound with $n$ in the denominator.
Part (2) Note that we have $|a_{n} - \frac{3}{2}| < \frac{3}{n}$ but really what we want is to show that for any $\varepsilon>0$ there is an $n$ such that $|a_{n} - \frac{3}{2}| < \varepsilon$. One way to show this is to ensure $n$ is large enough so that $\frac{3}{n}<\varepsilon$ (given me any $\varepsilon$ and I can pick an $n$ so this is true!). However, rearranging the inequality, choosing this $n$ is equivalent to choosing $n$ such that $n>\frac{3}{\varepsilon}$. 
Part (3) Suppose in (2) that we have found an $n$ so that this is true (i.e. that $\frac{3}{n} < \varepsilon$ for the chosen $\varepsilon$). Just for consistency of notation lets let this $n$ be $n_{0}$. Then $\frac{3}{n_{0}} < \varepsilon$.
However we already know that $|a_{n} - \frac{3}{2}| < \frac{3}{n}$ for any $n$, so it is also true for $n_{0}$. I.e. we have that $|a_{n_{0}} - \frac{3}{2}| < \frac{3}{n_{0}}$. Since $\frac{3}{n_{0}} < \varepsilon$ we have that $|a_{n_{0}} - \frac{3}{2}| < \varepsilon$.
The key is that I can do this for ANY $\varepsilon$. Given me any really really small number $\varepsilon$ and I can find an $n_{0}$ such that $|a_{n_{0}} - \frac{3}{2}| < \varepsilon$. BY DEFINITION of the limit (see your text or lecture notes), this implies that the limit of the sequence $\{a_{n}\}$ is $\frac{3}{2}$.
For your proof...
I will structure the proof in the same way as the first proof. 
Note that 
$| a_{n} - \frac{1}{3}| = |\frac{2n+5}{6n-3}-\frac{1}{3}| = \frac{18}{18n-9}\leq \frac{2}{n} < \frac{3}{n}$.
Thus for any $\varepsilon$ chose $n_{0}$ such that $n_{0}>\frac{3}{\varepsilon}$. Then we have that  $| a_{n_{0}} - \frac{1}{3}| < \varepsilon$. 
