<s>[INST] Is this method of showing that $\frac{2n^2}{n^3+3} \rightarrow 0$ correct? Given
$$
a_n = \frac{2n^2}{n^3+3}
$$
I want to show that
$$
a_n \rightarrow 0
$$
My solution proceeds as follows.
\begin{align}
\left\lvert \frac{2n^2}{n^3+3} - 0 \right\rvert &< \epsilon \\
\frac{2n^2}{n^3+3} &< \epsilon
\end{align}
since
\begin{align}
\frac{2n^2}{n^3+3} &< \frac{2n^2}{n^3} \\
&= \frac{2}{n}
\end{align}
then
\begin{align}
\frac{2}{n} < \epsilon \implies n > \frac{2}{\epsilon}
\end{align}
Therefore, let $N = \frac{2}{\epsilon}$. Fix $\epsilon > 0$ such that $\forall n > N$,
\begin{align}
\left\lvert \frac{2n^2}{n^3+3} - 0 \right\rvert < \frac{2}{n} < \frac{2}{N} = \epsilon
\end{align}
I am relying on the fact that since $\lvert a_n - 0 \rvert$ is upper-bounded by $\frac{2}{n}$, then the $N$ that I find for $\frac{2}{n}$ will be greater than the $N$ that I find for $\lvert a_n - 0 \rvert$.
Is all of this correct? If so, is there a name for this kind of method/reasoning?
 [/INST] Two points:

*

*You set $N:=\tfrac{2}{\varepsilon}$. But what is $\varepsilon$? Also $N$ need not be an integer. That's fine, but the symbol $N$ does suggest an integer value.

*Next you 'fix $\varepsilon$ such that $\forall n>N$...'.
But to prove that the limit equals $0$, you want to show that $$(\forall\varepsilon>0)(\exists N)\left(n>N\quad\Longrightarrow\quad \left|\tfrac{2n^2}{n^3+3}-0\right|<\varepsilon\right).$$
So instead, you want to start off from something like 'Let $\varepsilon>0$ be given...', and then give an appropriate value of $N$, and show that the inequality does indeed hold.
Your first few steps are a good way to determine such an $N$ given some $\varepsilon>0$. You can follow these steps in reverse to show that given $\varepsilon>0$, the desired inequality holds for all $n>N$ for the $N$ that you found.

For example, my answer to the question

Show that $\lim_{n\to\infty}\frac{2n^2}{n^3+3}=0$.

would simply be:

Given any $\varepsilon>0$, for all $n>\frac{2}{\varepsilon}$ we have
\begin{align}
\left\lvert \frac{2n^2}{n^3+3} - 0 \right\rvert <\left\lvert\frac{2n^2}{n^3}\right\rvert=\frac{2}{n}<\epsilon.
\end{align}

</s>