Proving the limit of $\frac{n+2}{n^3 -4n-2}$ using first principles. I'm trying to prove the limit of the sequence of $x_n=\frac{n+2}{n^3 -4n-2}$ using first principles. Here's what I've managed.
First I find the proposed limit using the algebra of limits.
$$x_n=\frac{n+2}{n^3 -4n-2} = \frac{\frac{1}{n^2}+\frac{2}{n^3}}{1-\frac{4}{n^2}-\frac{2}{n^3}}$$
Therefore as $n \rightarrow \infty$ the limit is proposed to be $0$. 
Fix $\epsilon >0 $ and $N \in \mathbb{N}$,
$$ n\geq N \implies \bigg| \frac{n+2}{n^3 -4n-2} -0\bigg| < \epsilon$$ 
For all $n \geq 3$.
$$\bigg| \frac{n+2}{n^3 -4n-2} -0\bigg| = \frac{n+2}{n^3 -4n-2}$$ 
$$\frac{n+2}{n^3 -4n-2} \leq \frac{2}{n} < \epsilon \iff  \frac{2}{\epsilon} < n $$
Therefore choose $N$ greater than $Max\{3,\frac{2}{\epsilon}\}$, has the required property.
Any feedback would be much appreciated 
 A: Your proof is fine, except you have not provided details of how $\frac{n+2}{n^3-4n-2} \leq \frac 2n$ for $n \geq 3$.
How would I prove this? Cross multiply first, to see that it is enough to prove $n^2+2n \leq 2n^3-8n -4$ for $n \geq 3$. At $n=3$, this inequality is true by checking, and if it is true for $n=k$, then when I move to $k+1$ the LHS increases by $2k+3$ while the RHS increases by $6k^2+6k+2 - 16$, which by usual theory for quadratics is seen to be bigger than $2k-2$ for $k \geq 3$. 

For an alternate proof, using the heuristic, you would notice :
$$
\frac{\frac 1{n^2} + \frac 2{n^3}}{1 - \frac 4{n^2} - \frac 2{n^3}} = \frac {a_n}{b_n}
$$
where $a_n$ goes to $0$ monotonically and $b_n$ goes to $1$ monotonically as $n \to \infty$. Therefore, you "mock" the proof of quotient of limits as follows : for $\epsilon > 0$, choose $N$ large enough so that $a_n < \frac \epsilon 2$ for all $n > N$, and $b_n > \frac 12$ for all $n > N$ (by taking a maximum). The quotient is then at most $\epsilon$ (it has to be positive for such large $N$, so must lie in $(0,\epsilon)$) . Conclude.
