Epsilon limit question proof verification/help (self study) I have to show the following $\lim \frac{2n^2}{n^3+3}=0$.
So I have to show that $\forall \epsilon>0,  \exists N \in \mathbb{N} :\ when\ n \geq N$ it follows that $|a_{n}-a|<\epsilon$.
I choose $\epsilon= \frac{1}{5}$, thus $a_{n} \in ( \frac{1}{-5}, \frac{1}{5})$ when $n>3$ so I pick $N=4$.
So I have to show that $\forall \epsilon, \exists N : |\frac{2n^2}{n^3+3}-0|< \epsilon$, 
and then I am not sure how to continue. I have shown that for which $N$ will the sequence stay inside the epsilon interval. But turning it into a rigorous proof is a little bit hard to imagine, since this is my first proof with this technique.
 A: In order to have rigor, we want to algebraically show that there exists an $N$ for each $\epsilon$. What you are doing right now looks a bit like trial and error, so we need to find some trick that works well. Here is a trick (which Jair Taylor also pointed out) that turns out to be very helpful here:
$$\frac{2n^2}{n^3+3}<\frac{2n^2}{n^3}=\frac{2}{n}$$
Now if we pick an $\epsilon$, then we can find an $N$ which works, since we want $\frac{2}{N}<\epsilon$, we can rearrange as $N>\frac{2}{\epsilon}$, so just take $N=\lfloor\frac{2}{\epsilon}\rfloor + 1$, and we're done!
A: Notice: 
$$\frac{2n^2}{n^3 + 3} < \frac{2n^2}{n^3} = \frac{2}{n}$$
Now, let $\epsilon > 0$ be given. Choose $N$ so large that $2/N < \epsilon$, which is possible by the Archimedian property of the real numbers.
Then, if $N \leq n$, it follows that $1/n \leq 1/N$, and hence $$\frac{2n^2}{n^3 + 3} < \frac{2}{n} \leq \frac{2}{N} <\epsilon$$
Since $\epsilon$ was arbitrary, we have proven that:
$$\forall \epsilon > 0: \exists N : \forall n \geq N: \frac{2n^2}{n^3 + 3} < \epsilon$$
which means that $$\lim_{n \to \infty} \frac{2n^2}{n^3 + 3} = 0$$
as desired.

Notice also that we could have immediately applied the squeeze/sandwich theorem. Indeed, we have:
$$0 \leq \frac{2n^2}{n^3 + 3} \leq \frac{2}{n}$$
and both the left and the right side go to $0$ if $n \to \infty$, meaning that the middle expression goes to $0$ as well.
A: $a_n:=\dfrac{2n^2}{n^3+3} \lt \dfrac{2n^2}{n^3} \lt \dfrac{2}{n}.$
Let $\epsilon >0$ be given.
Let $M:= 2/\epsilon$.
Archimedean principle:
There is a $n_0 > M$, $n_0 \in \mathbb{Z^+}$.
For $n \ge n_0$:
$|a_n-0| = \dfrac{2n^2}{n^3+3} \lt \dfrac{2}{n} \le \dfrac{2}{n_0} \le $
$\dfrac{2}{M} = \epsilon.$
