Verifying convergence of sequences in exercise 2.2.2 of Understanding Analysis by Stephen Abbott (2015). I am having difficulty with what should be a routine question, Exercise 2.2.2 (b) of Understanding Analysis by Stephen Abbott (2015).

Exercise 2.2.2. Verify, using the definition of convergence of a sequence, that the following sequences converge to the proposed limit.
(b) $$\lim_{n \rightarrow \infty} \frac{2n^2}{n^3 + 3} = 0.$$

My attempt.
Due to a transcription error in my question and some answers based on my erroneous transcription, followed by re-edits etc., this has caused some confusion. To avoid any further future confusion, I have reverted this question, together with my attempt to the most recent state before I received correct answers, and upvoted answers accordingly. For the members of the community who took the time to assist, please accept my apologies, and also my gratitude for the time you've spent writing up your answers.
I understand that verification using the formal definition (without recourse to theorems occurring later in the text) requires supplying a particular $N(\epsilon) \in \mathbb{N}$ in response to any $\epsilon > 0$ such that whenever $n \geq N$, then
$$\left \vert \frac{2n^2}{n^3+3} \space \right \vert < \epsilon.$$
Because $n > 0$, I can simplify the above to get
$$\left \vert \frac{2n^2}{n^3+3} \space \right \vert = \frac{2n^2}{n^3 + 3} < \epsilon.$$
Which yields the following cubic polynomical in $n$, which I am unsure how to solve analytically:
$$\epsilon n^3 - 2n^2 + 3 \epsilon > 0.$$
 A: Let $\frac{1}{2}\varepsilon>\frac{1}{n}>0$ by the Archimedean property (I believe this is defined in Abbott's book).
Now, observe the following for $n\in \mathbb{N}$:
$\Big|\frac{2n^2}{n^3+3}-0\Big|=\Big| \frac{2n^2}{n^3+3} \Big|=\frac{2n^2}{n^3+3}<\frac{2n^2}{n^3}=\frac{2}{n}<2\cdot\frac{1}{2}\varepsilon=\varepsilon$
and thus we have proven the limit to be zero.
Edit: I edited the answer after you gave the correction.
A: Alternative
Rewrite $~~~\frac{2n^2}{n^3 + 3}~~~$ as $~~~\displaystyle \frac{2}{n + \frac{3}{n^2}}.$
So, the problem reduces to showing that as 
$\displaystyle n \to \infty, ~\left(\frac{2}{n + \frac{3}{n^2}}\right) \to (0)$.
Since, for all $(n)$, you have that $\displaystyle \left(n + \frac{3}{n^2}\right)$ is strictly greater than $(n)$, 
the desired conclusion may be reached from the intermediate result that 
$\displaystyle \lim_{n\to\infty} \frac{2}{n} = 0.$
A: After the edit to the OP which changes the $n^2$ in the denominator to $n^3$:
Observe that
$$2n^3 \leq 2n^3 + 6 = 2(n^3 + 3)$$
so
$$2n^2 = \frac{2n^3}{n} \leq \frac{2(n^3+3)}{n}$$
Dividing the left and right sides by $n^3 + 3$ gives
$$\frac{2n^2}{n^3 + 3} \leq \frac{2}{n}$$
Thus the LHS is sandwiched between $0$ and $2/n$, and the latter converges to $0$ as $n \to \infty$.
A: As mentioned in the comments, based on the way the question is written, the limit should be $2$ and not $0$.
Assuming the question is correct as written, we have
$$\left| \frac{2n^2}{n^2+3} - 2 \right| < \epsilon \implies \left| \frac{2n^2-2(n^2+3)}{n^2+3} \right| < \epsilon \implies \left|\frac{-6}{n^2+3} \right|<\epsilon$$
Since $n^2+3$ will be positive, this gives us
$$\frac{6}{n^2+3} < \epsilon$$
I'm sure you can take the rest from here.
