Q:

Verify Using the Definition of Convergence of a sequence, that the following sequences converge to the proposed limit.

(a) lim $\displaystyle\frac{2n^2}{n^3+3} = 0$
(b) lim $\displaystyle\frac{\text{sin}(n^2)}{n^{\frac{1}{3}}} = 0$
For (a) I have:
$$\bigg|\displaystyle\frac{2n^2}{n^3+3} - 0\bigg| < \epsilon$$
$$n^3+3> \displaystyle\frac{\epsilon}{2n^2}$$
$$n^3 > \displaystyle\frac{\epsilon}{2n^2} - 3$$
$$n> (\displaystyle\frac{\epsilon}{2n^2}-3)^{\displaystyle\frac{1}{3}}$$
Let $\epsilon>0$ be arbitrary. The inequality $\bigg|\displaystyle\frac{2n^2}{n^3+3} - 0\bigg| = \frac{2n^2}{n^3+3} < \epsilon$ is the same as $n> (\displaystyle\frac{\epsilon}{2n^2} -3)^{\displaystyle\frac{1}{3}}$. Therefore, $\forall \epsilon >0$, $\exists N \in \mathbb{N}$ such that $N> (\displaystyle\frac{\epsilon}{2n^2} -3)^{\displaystyle\frac{1}{3}}$. Then $\forall n \geq N$, it follows that $\displaystyle\frac{2n^2}{n^3+3} \leq \frac{2N^2}{N^3+3} < \epsilon$. Thus verifying the limit. 

A:

Let $\varepsilon>0$ be arbitrary. Then see that exists $N > \frac{2}{\varepsilon}$ because no matter how big the number $2/\varepsilon$ can be we have that $\mathbb{R}$ is an Archimedean ordered field so there must be a natural number larger then $2/\varepsilon$. So this implies that  for all $n > N$ we will have
$$n > \frac{2}{\varepsilon}  \implies n + \frac{3}{n^2}> \frac{2}{\varepsilon}$$
so
$$n + \frac{3}{n^2}> \frac{2}{\varepsilon} \implies \frac{2}{n + \frac{3}{n^2}}<\varepsilon \implies \frac{2n^2}{n^3+3}< \varepsilon$$
So we proved that, given an $\varepsilon > 0$ we always can find a natural number $N \in \mathbb{N}$ such that for all $n > N$ we get that 
$$\left\vert\frac{2n^2}{n^3+3}-0\right\vert < \varepsilon$$
For the second item the hint given above should help you.  

