Proving some sequence tends to zero using first principles. I'm taking a Real Analysis course at university and I having trouble understanding how to prove a sequence converges to zero as a natural number n tends to infinity. Here's the question.

I understand you have to fix in epsilon strictly greater than zero and find a N in the natural numbers such that.

I understand for all n equal to or greater than 3 the absolute value drops away. But here I get confused. The next step we have to replace the expression with one that is easier to deal with here's what the model solution used as its replacement for the top of the fraction.

I have two questions. The motivation to use 4n was to increase the top of the fraction but why did the choose 4n, why not 5n or 42n... Secondly why did they want to increase the top, I've seen problems where they increase the bottom of the fraction and decrease the top. Another Problem comes later when they replace the bottom of the fraction.

Why not a third or a quarter n^2. Why this very specific expression.
If someone could push me in the right direction I would really appreciate it.
 A: 
but why did the choose $4n$, why not $5n$ or $42n$ ...
Why not a third or a quarter $n^2$. Why this very specific expression.

It would work with those as well, for example if you show that the expression is $\leq \frac{42n}{\frac{1}{4}n^2}=\frac{168}{n}$, then still right side is easily shown to be arbitrary small. Reason they choose those numbers is just to make it simpler, that's all. Just note that this inequality is not true for all $n$, it is true for sufficiently large $n$, but since you are dealing with limit $n \to \infty$, you can throw away how many finite terms you need since that won't affect the limit.

Secondly why did they want to increase the top ...

Because they eventually want to show that the expression is $< \varepsilon$, by bounding the numerator from above and bounding the denominator from below, they effectively bound the fraction from above. So they are able to show $\frac{n+3}{n^2-2n-2} \leq \frac{4n}{\frac{1}{2}n^2}=\frac{8}{n}$, and then it is much easier to show that $\frac{8}{n}<\varepsilon$ for sufficiently large $n$, which by transitivity of $\leq $ means $\frac{n+3}{n^2-2n-2} < \varepsilon$, which was to be shown.
If you would need to show that a fraction $p/q \geq$ something, then you would do it the other way around, i.e. bounding $p$ from below and $q$ from above.
A: In analysis you can use whatever bounds you want as long as you can solve the problem. In your two examples, with enough experience you will be able to see that it doesn't really matter much of the time as long as you can see that eventually for some $n$ the inequality will be satisfied. Eventually for some $n$, both of those expressions will be satisfied so you can take the max of the two. Furthermore, you are left with the initial expression being less than $8/n$. Using the archimedian property we can make $n$ large enough so that $1/n < \varepsilon/8$. Take the max of these $3$ values and set it equal to $N$. Taking the max allows the line of logic to be reversed, as is customary in solving analysis problems.
A: $4n$ is because for all $n\ge1,4n\ge n+3$. Any factor larger than $4$ could have been taken.
If fact, $2n$ would have been good enough because $2n\ge n+3$ for all $n\ge 3$, and we don't really care about the low values of $n$. (Even $1.000001n$ could do.)
But you know how mathematicians are, perfectionists.
