<s>[INST] Proving that $a_n=\frac{n^2+2n+6}{n^3-3}\to 0$ as $n\to\infty$ $$a_n=\frac{n^2+2n+6}{n^3-3}$$
So I want to show that "$a_n\to a\iff\forall \epsilon>0,\quad\exists N\in\mathbb{N}:n\geq N\implies |a_n-a|<\epsilon$"
Then my rough working:
$|a_n-0| =\left|\frac{n^2+2n+6}{n^3-3}\right|<\epsilon$
Estimate $\frac{n^2+2n+6}{n^3-3}<\frac{b_n}{c_n}$
Require $b_n>n^2+2n+6$, so choose $b_n=2n^2>n^2+2n+6$ for $n\geq 4$
Require $c_n<n^3-3$, so choose $c_n=\frac{n^3}{2}<n^3-3$ for $n\geq 2$
Then let $N=\frac{2n^3}{n^3/2}=\frac{4}{n}$
So now the proof:
$$\text{Fix}\quad \epsilon>0.\quad\text{Pick}\quad N:N>\frac{4}{n}\quad\text{and}\quad N\geq 4$$
$$\implies \forall n\geq N,\quad |a_n-0|=\left|\frac{n^2+2n+6}{n^3-3}\right|<\frac{4}{n}\leq\frac{4}{N}\leq\epsilon$$
Is this correct? Is it completely rigorous? I've just started doing analysis on my own, so please point out if anything is wrong/not conventional so that I can get into good habits at the start. Thanks for any help.
 [/INST] If i were you, i would just multiplicate numerator and denominator by $1/n^2$ both, so that: $a_n = (1 + 2/n + 6/n^2)/(n -3/n^2)$. From the defenition of infinite number, we have $2/n$, $6/n^2$  and $-3/n^2$ all equal to 0: $a_n = (1 + 0 + 0)/(n + 0 + 0) = 1/n$ which is 0 from the same defenition.
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