Find $\lim _{r\to \infty}\frac{\left(\prod_{n=1}^{r}\sin\left(nx\right)\right)}{\left(\frac{1}{r}\right)}$ $\lim _{r\to \infty}\frac{\left(\prod_{n=1}^{r}\sin\left(nx\right)\right)}{\left(\frac{1}{r}\right)}$ 
I tried using the product sin formula but got nowhere.& even after multipling and dividing by $2cos (x)$,   answer couldnt be obtained as only multiples of two cut out themselves.
it is a 0/0 indeterminate form.
Also I didn't get the answer by using the l's-hopital rule.
I even tried graphing it on desmos, but the graph was strange---( I think even desmos couldn't compute it further)
please help.
thanks in advanced..............
 A: If $x$ is a rational multiple of $\pi$, then for some integer $N > 0$, $\sin(Nx) = 0$. This forces $\prod_{n=1}^r \sin(nx) = 0$ whenever $r \ge N$. In this case, the limit is $0$.
Otherwise, $x$ is not a rational multiple of $\pi$ and $|\cos x| < 1$. Notice
$$|\sin(nx)\sin(n+1)x| = \frac{|\cos x - \cos((2n+1)x)|}{2} \le \mu \stackrel{def}{=}\frac{1 + |\cos x|}{2}$$
By grouping the factors in the numerator in pairs, we have following bound
for the weighted product at finite $r$.
$$r\left|\prod_{n=1}^r \sin(nx)\right|
\le r\prod_{k=1}^{\lfloor r/2\rfloor} |\sin((2k-1)x)\sin(2kx)|
\le r\mu^{\lfloor r/2\rfloor}
$$
Since $\mu < 1$, the limit of the weighted product is again $0$.
A: If $x=\frac pq \pi$ for $p,q\in \mathbb Z, (q\ne 0)$, then the limit would evaluate to zero as for any $q$ we have the term $\sin(qx)$ in our product. If $x$ is not a rational multiple of $\pi$, then no term in the numerator equals zero and we have to take the limit. 
$\because -1 \lt \sin(mx) \lt 1$ doesn’t vary much, the limit can be reasonably approximated as:
$$\lim_{r\to\infty} r\prod_{n=1}^r \sin(nx) \sim \lim_{r\to\infty} r (\sin x)^r \\ =\lim_{r\to\infty} \frac{r}{(\csc x)^r} =0$$
$ \because|\csc x| \gt 1$, the limit equals zero due to the fact exponentials grow way faster. 
