How can I find the limit of a sequence containing $\sin n!$? $$\begin{equation*}
\lim_{n \rightarrow \infty}
\frac{n^{2/3} \sin n!}{n+1}
\end{equation*}$$
I know that I should divide numerator and denominator by the highest power of n in the denominator but then what? how can I deal with the $n!$ which is the angle of the sine? I know that $n!$ is larger than $n$ and I know also that $|\sin n| \leq 1$, but how can I organize all this information to use it in my solution?
Could anyone help me please?   
 A: You don't have to worry about the factorial at all, since $|\sin(n!)|\le1$ just like $|\sin n|$.  So
$$\left|n^{2/3}\sin(n!)\over n+1 \right|\le\left|n^{2/3}\over n+1 \right|\le\left| n^{2/3}\over n\right|={1\over n^{1/3}}\to0$$
A: You have
$$0\leq |\frac{n^{2/3}\sin n!}{n+1}|=\frac{n^{2/3}|\sin n!|}{n+1}\leq\frac{n^{2/3}}{n+1}\leq\frac{n^{2/3}}{n}=n^{-1/3}\to 0(n\to\infty)$$
To elaborate, the first two steps are just rules for absolute values and products. Note that $|\sin n!|$ ranges between $0$ and $1$, as $\sin n!$ ranges between $-1$ and $1$, i.e. as an upper bound $|\sin n!|\leq 1$. The factorial has no impact as the sine, ranging from $-1$ to $1$, is periodically over all of $\mathbb{R}$. Lastly, we just decrease the denominator to finally cross out the $n$ term.
By the Squeeze theorem, it follows that $$\lim_{n\to\infty}|\frac{n^{2/3}\sin n!}{n+1}|=0$$ and as $\lim_{n\to\infty}a_n=0$ iff $\lim_{n\to\infty}|a_n|=0$ for any real sequence $(a_n)_n$, we have $$\lim_{n\to\infty}\frac{n^{2/3}\sin n!}{n+1}=0$$
