Our sequence is $\dfrac{\sin (n^2)}{\sqrt[3]{n}}$.
Now, what we are going to do first, is the following:
note that $\dfrac{-1}{\sqrt[3]{n}} \leq \dfrac{\sin (n^2)}{\sqrt[3]{n}} \leq \dfrac{1}{\sqrt[3]{n}}$. Rewritten succinctly, $\Bigg|\dfrac{\sin (n^2)}{\sqrt[3]{n}}\Bigg| \leq \Bigg|\dfrac{1}{\sqrt[3]{n}}\Bigg|$
Let $\epsilon > 0$. Now, there exists $N \in \mathbb{N}$ such that $n > N \implies \Bigg|\dfrac{1}{\sqrt[3]{n}}\Bigg| < \epsilon$. Namely, we can let $N = \dfrac{1}{\epsilon^3}$, for example.
Therefore,
$$
n > N \implies \Bigg|\dfrac{\sin (n^2)}{\sqrt[3]{n}}-0\Bigg| \leq \Bigg|\dfrac{1}{\sqrt[3]{n}}\Bigg| < \epsilon
$$
Hence, we are done, with $a=0$.
What was important in this proof, was the use of the boundedness of the $\sin$ function. It allowed us to considerably simplify the calculation of $N$ given $\epsilon$.
Best Answer
Let $\epsilon>0$. $\sin(x)$ is continueous at $x=0$ and $\sin(0)=0$.
As a result there exists $a>0$ such that $\forall x\in[-a,a],|\sin(x)|<\epsilon$.
What is more since $\lim_\infty\frac{1}n=0$ we have $N>0$ such that $\forall n\ge N,\left|\frac{1}n\right|<a$.
As a result $\forall n>N,|\sin(\frac{1}n)|<\epsilon$.
Conclusion : $\boxed{\lim_{n\to\infty}\sin(\frac{1}n)=0}$