The limit of $f_n = \frac {e^x \sin(x) \sin(2x) \cdots \sin(nx)}{\sqrt n}$ 
I want to calculate a limit of $f_n = \frac {e^x \sin(x) \sin(2x) \cdots
 \sin(nx)}{\sqrt n}$ if $n$ goes to an infinity.

I was wondering if I can use this: $ \lim_{n\to \infty} \frac {\sin(nx)}{n} = 0$, because I have doubts.
I know that $$  \lim_{x\to x_0} f(x)g(x) =  \lim_{x\to x_0}g(x)\lim_{x \to x_0} f(x)$$ (or I suppose), but what if the limit of $f(x)$ or $g(x)$ is zero
My attempt of a solution:
$$ \lim_{n\to \infty} \frac {e^x \sin(x) \sin(2x) ...
\sin(nx)}{\sqrt n} = \lim_{n\to \infty} \frac {\sqrt ne^x \sin(x) \sin(2x) ...
\sin(nx)}{ n}\\ = \lim_{n\to \infty}  {\sqrt ne^x \sin(x) \sin(2x)}  \frac {\sin(nx)}{n} = \lim_{n\to \infty}  {\sqrt ne^x \sin(x) \sin(2x)} * 0 = 0$$
And the range for the convergence is $\infty$ , right?

For a given $x$, we can show that $\inf \{ m\cdot x - n\cdot\pi: m, n \in {\mathbb Z}\} = 0$.  Thus, $|\sin mx|$ can be made arbitrarily small by proper choice of $m$. The other sine factors and the $1/\sqrt n $ factor reduce the modulus, and the $e^x$ factor is fixed.  The limit is, therefore, 0.
