Prove $\lim\frac{4\sin(n^2)}{3n}=0$ 
Prove $\lim\frac{4\sin(n^2)}{3n}=0$

Using the fact that $\left|s_n-s\right|\lt \epsilon$
I'm finding it difficult to solve for $n$.  I recognize the function is bounded between $-1$ and $1$, but I don't know how to use this information to help with my proof.  
I was able to get it into the below form:
$\sin(n^2)\lt \frac{3n\epsilon}{4}$
But I'm not sure where to go from here.
 A: I'm assuming the limit is $n \to \infty$, so we want to show that, for any $\epsilon > 0$, there is some $N$ such that:
$$n > N \implies \left\lvert \frac{4\sin(n^2)}{3n} \right\rvert < \epsilon$$
Again, we're given $\epsilon$ and we want to find $N$. I think you were doing something like this, but I just want to clarify.
Now, we can do the following with this expression:
$$\left\lvert \frac{4\sin(n^2)}{3n} \right\rvert=\frac{\left\lvert 4\sin(n^2)\right\rvert}{\left\lvert 3n\right\rvert}=\frac{4\lvert\sin(n^2)\rvert}{3\lvert n\rvert}$$
We know that $\lvert\sin(n^2)\rvert \leq 1$ and since $n \to \infty$, we can assume $n > 0$ so $\lvert n \rvert=n$, so we get:
$$\left\lvert \frac{4\sin(n^2)}{3n} \right\rvert=\frac{4\lvert\sin(n^2)\rvert}{3\lvert n\rvert} \leq \frac{4}{3n} < \epsilon$$
Solving for $n$, we get $n > \frac{4}{3\epsilon}$. Thus, if we are given $\epsilon$, we can choose $N=\frac{4}{3\epsilon}$ to satisfy the original conditional at the top, concluding the proof.
A: Hint:
For any real number $t$ we have $-1\leq \sin t\leq 1$. Now, given $\varepsilon>0$ take $n>\frac{4}{3\varepsilon}$, so $$-\frac{4}{3\left(\frac{4}{3\varepsilon}\right)}<-\frac{4}{3n}\leq \frac{4\sin(n^2)}{3n}\le\frac{4}{3n}<\frac{4}{3\left(\frac{4}{3\varepsilon}\right)} \qquad\implies\qquad -\varepsilon<\frac{4\sin(n^2)}{3n}<\varepsilon$$
A: Well: $|sin(n^2)|\leq1$
So:
$|\frac{sin(n^2)}{n}|\leq \frac{1}{n}$
Hopefully you can finish from here.
A: I am opting to use Squeeze theorem. 
It is known that 
$-1 \le \sin({n^2}) \le 1$
$\rightarrow -\frac{4}{3n} \le \frac{4\sin({n^2})}{3n} \le \frac{4}{3n}$
$\lim_{n\rightarrow \infty}(-\frac{4}{3n})=\lim_{n\rightarrow \infty}(\frac{4}{3n})=0$
$\rightarrow \lim_{n\rightarrow \infty} \frac{4\sin({n^2})}{3n}=0$ (By squeeze theorem)
A: Use the squeeze theorem: 4/(3n) goes to 0 and so does -4/(3n). Our function is between these functions (in value at each point) so it goes to 0.
