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.
 A: 1. You just bound in by making the denominator smaller and numerator bigger, this is a common trick in these proofs.
$$\frac{11}{2(2n+7)}<\frac{11}{2(2n)}=\frac{11}{4n}<\frac{12}{4n}=\frac3n$$
These proofs have TWO STEPS. Basically, you want to do the "prep work", then the proof. Let me write out this proof in full. The above is the PREP WORK and not actually the proof.
Proof: Let $\epsilon >0$ be given. By the Archimedian property of the natural numbers, there is some $n_0>\frac{3}{\epsilon}$. Thus if $n>n_0$, then $n>\frac{3}{\epsilon}$. Rearranging gives that $\epsilon>\frac{3}{n}$. Thus:
$$\epsilon>\frac{3}{n}=\frac{12}{4n}>\frac{11}{4n}>\frac{11}{2(2n+7)}=\left|a_n-\frac{3}{2}\right|$$
That is, for all $\epsilon>0$ there is some $n_0$ such that for all $n>n_0$. $|a_n-\frac{3}{2}|<\epsilon$. This is the definition of limit, so $\lim\limits_{n\to\infty} a_n=\frac{3}{2}$.
A: 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
$$n\ge n_0\quad\Longrightarrow\quad\left|a_n-\frac{1}{3}\right|=\left|\frac{2}{2n-1}\right|<\left|\frac{2}{2n-2}\right|=\frac{1}{n-1}<\frac{1}{1/\epsilon}=\epsilon$$
A: 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  $| a_{n_{0}} - \frac{1}{3}| < \varepsilon$. 
A: I'll run through your questions.


*

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


It is an upper bound on $\left|a_n-\frac{3}{2}\right|=\frac{11}{2(2n+7)}$, and one way it can pop up is as follows:
$$\frac{11}{2(2n+7)}=\frac{11}{4n+14}<\frac{11}{4n}<\frac{12}{4n}=\frac{3}{n}.$$
It is meant as some simple, vanishing upper bound of the difference $\left|a_n-\frac{3}{2}\right|$.


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


What happens here is that we choose some large $n_0\in\mathbb{N}$, such that $n_0 > \frac{3}{\epsilon}$. Then certainly $\frac{3}{n_0}<\epsilon$, and more importantly, $\frac{3}{n}<\epsilon$ for all $n\geq n_0$. Thus, since we have that
$\left|a_n-\frac{3}{2}\right|<\frac{3}{n},$ we have also
$$\left|a_n-\frac{3}{2}\right|<\epsilon,$$
for all $n\geq n_0$. So it's not that we replace $n$ by $\epsilon$, but that we can choose $n$ so large that $\frac{3}{n}$ is less than the $\epsilon$ given.


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


This follows from the definition of limits. No matter what $\epsilon>0$ I start out with, the limit of $a_n$ will be $\frac{3}{2}$, if
$$\left|a_n-\frac{3}{2}\right|<\epsilon$$
for large enough $n$ (in our case, for $n\geq n_0$. This is the same as $\lim_{n\rightarrow\infty} a_n=\frac{3}{2}$.
