Using the definition of the convergence of the sequence, how can I prove that this sequence converges to the limit? I suppose what I am confused most on here is the algebraic method.  
$\lim_{x\to\infty} $ $ 2n^2/(n^3 + 3) $ = 0
I have set it up so far: 
Let $ \epsilon > 0 $ be given. 
I set up the equation as follows, it's just the algebra I'm getting stuck on:
$ |2n^2/(n^3 + 3) - 0| < \epsilon $
$ |2n^2/(n^3 + 3)| < \epsilon $
$2n^2/(n^3 + 3)< \epsilon $
$ 2n^2 < n^3\epsilon + 3\epsilon $
$ 2n^2 - n^3\epsilon < 3\epsilon $
$ n^2(2 - n\epsilon) < 3\epsilon $
If I could isolate n and replace it with N, I would be able to complete this proof. Does anyone have the solution to this problem in terms of set up? Thank you!
Also, similarly with the algebra, could anyone help me with this proof as well? 
$\lim_{x\to\infty} $ $ sin(n^2)/ \sqrt[3]{n} $ = 0
 A: By multiplying the fraction by $\frac{1/n^2}{1/n^2}$, this sequence becomes 
$$\frac{2}{(n+\frac{3}{n^2})}$$
From this point, it should be pretty clear that the numerator is fixed and the denominator goes to $\infty$, so this must go to $0$. To explicitly solve for $N$, observe that this is bounded above by 
$$\frac{2}{n+\frac{3}{n^2}} \leq \frac{2}{n},$$
so for any $\epsilon>0$ if $n\geq N$, where $N=2/\epsilon$, then it must be that
$$\frac{2}{n+\frac{3}{n^2}}\leq \frac{2}{n}\leq \frac{2}{\frac{2}{\epsilon}}=\epsilon.$$
where the last inequality comes from $\frac{2}{n}$ is decreasing as $n$ gets larger. 
Similar logic can be used for your question about $\frac{sin(n^2)}{\sqrt[3]{n}}$. This is bounded above by $|\frac{sin(n^2)}{\sqrt[3]{n}}|\leq \frac{1}{\sqrt[3]{n}}$, which can then be used to solve for $N$.
A: Perhaps you would like to consider the following: the inequality $$\frac{2n^2}{n^3 + 3}< \frac{2n^2}{n^3}=\frac{2}{n}.$$ In this case, if $N$ is an integer larger than or equal to $\frac{2}{\epsilon}$, the sequence would converge to $0$.
