Proving a sequence has limit using an epsilon - N argument I have two separate sequences that I using this approach on.
$a_n=\frac{n^3-2n^2+3}{2n^3+7n}$ & $a_n=\frac{n^3}{2^n}$
Proof 1
$a_n=\frac{n^3-2n^2+3}{2n^3+7n}\to \frac{1}{2}$ 
$$\text{Let } \epsilon \gt 0 \,\, \exists N \;\;\forall n\geq N \; \left\lvert {a_n-\frac{1}{2}}\right\rvert \lt \epsilon$$
$$\left\lvert {\frac{n^3-2n^2+3}{2n^3+7n}-\frac{1}{2}}\right\rvert=\left\lvert {\frac{(2n^3-4n^2+6)-(2n^3+7n)}{4n^3+14n}}\right\rvert=\left\lvert {\frac{-4n^2-7n+6}{4n^3+14n}}\right\rvert \lt \frac{1}{2n} \lt \epsilon$$
$$\frac{1}{2\epsilon} \lt n$$
I am having difficulty in showing exactly why, $\left\lvert {\frac{-4n^2-7n+6}{4n^3+14n}}\right\rvert \lt \frac{1}{2n}$
Proof 2
$a_n=\frac{n^3}{2^n} \to 0$
$$\text{Let } \epsilon \gt 0 \,\, \exists N \;\;\forall n\geq N \; \left\lvert {a_n-0}\right\rvert \lt \epsilon$$
$$\left\lvert {\frac{n^3}{2^n}}\right\rvert=\frac{n^3}{2^n} \lt  \frac{?}{?} \lt \epsilon$$
 I am lost as to where to proceed on this one. Any input would be greatly appreciated . Thanks

On the first proof:
It would be simplest to show that $\left|\dfrac{4n^2+1}{4n^3+14n}\right|<\dfrac{1}{n}$ since, clearly $4n^3+n<4n^3+14n$.
On the second proof:
For $n\ge16$ we have that $n^4\le2^n$ from which it follows that $\dfrac{n^3}{2^n}\le\dfrac{1}{n}$.
