Proof a sequence converges to a limit For a sequence
$$a_n = \frac{\sin(n)+2}{4n^2-28}$$
How would you use the definition of a limit of a sequence to prove $a_n$ converges to $0$ 
I am really stuck with how this definition works, I understand you need to find a value for $N>$ but all examples end up with a min/max answer and I have no idea how they got there. 
Why is the answer $\min\{5,1/\sqrt{\varepsilon}\}$?
When I look online I can only find examples using the definition of convergence of a function and not sequence.
What are the rules for cancelling down when and how and why do you get min/max  answers?
Really looking for a step by step answer thank you.
 A: $$|a_n - l |= \left|{\frac{\sin(n)+2}{4n^2-28}}\right|  = \frac{|\sin(n)+2|}{4|n-\sqrt 7||n + \sqrt 7|}$$
Now $n\ge 4 \implies n - \sqrt 7 \gt 1$. Therefore, 
$$|a_n - l | \lt  \frac{|\sin(n)+2|}{4|n + \sqrt 7|} \le  \frac{|\sin(n)| +2}{4(n + \sqrt 7)}    \le  \frac{3}{4n }$$
Now let $\epsilon \gt 0$ be an arbitrary natural number.  Since $\Bbb N$ is unbounded there must be a natural number, $N$, which is greater than $\text{Min} \{ \frac{3}{4\epsilon}, 4 \}$. Then,
$$n \ge N \implies |a_n - l| \lt \frac{3}{4n } \le \frac{3}{4N }  \lt \epsilon \implies \lim(a_n) = l = 0 $$  

I thought of explaining all of this a little bit more. When you are asked to prove the convergence of a sequence what you need to do is to make the quantity $|a_n - l|$ as small as possible($l$ is the limit here). The best way to do this is to show that it is smaller than $\frac{A}{n}$ where $A \gt 0$ and $n$ is a positive number. This quantity can be made as small as you wish by choosing $n$ as large as required. To do this you need to use inequalities. What we have done here is show that the denominator and numerator of $|a_n - l|$ can be approximated using other fractions. For an example if $|4n^2 - 28| \gt n^2 $, then $ \frac{|\sin n + 2|}{|4n^2 - 28|} \lt \frac{|\sin n + 2|}{n^2 } $. And you use similar steps to get to a point where you can say that if $n$ is large enough then |$a_n - l|$ is as small you want i.e. is less than any $\epsilon \gt 0$, however small. Now, according to your answer if we choose $n$ such that $n \gt \min\{5,1/\sqrt{\varepsilon}\} $ then you must see that $ \frac{1}{n^2} \lt \epsilon \; $ and $4n^2 - 28 \gt n^2$ satissfying both our conditions. And we are done. 
A: Well, it seems pretty straightforward to me, so I wonder if I am missing something.  Wouldn't be the first time.  But anyway ...
$\sin(n)$ can never be greater than 1 or less than -1.  Therefore the numerator, whatever its value for any particular $n$, is always less than or equal to 3 (i.e. 1+2) and greater than or equal to 1 (i.e. -1+2).  The sequence $3/(4n^2 - 28)$ and the sequence $1/(4n^2-28)$ both converge to zero for $n\rightarrow\infty$, because the denominator grows toward infinity.  The limit of your given sequence must lie in between these two limits, i.e. between zero and zero, and must therefore go to zero.
A: Fix $\epsilon >0$. Now you should try to find a positive integer $N$ such that, whenever $n >N$, you get $|a_n|<0$. Since you already know the answer, you should deduce that the sequence shows some decreasing pattern for $n>5$ (this probably comes from the factor $4n^2-28$ which is positive for $n\geq 5$, $n$ integer positive.
Also, note that $|sin(n)|\leq 1$ for all $n$. I would also try using triangle inequality on $|sin(n)+2|$. The standard approach here is to compare $|a_n|$ with some arbitrary positive $\epsilon$ and get some information on $\epsilon$.
