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?
 A: Hint: $\sin(x)$ is always between 1 and 0. But the denominator keeps getting larger as n goes to infinity.
A: The pointwise convergence towards zero is trivial, we may actually prove we have uniform convergence over any compact subset of $\mathbb{R}$.
Let
$$ g_n(x) = \sin(x)\sin(2x)\cdots \sin(nx). $$
The supremum of $g_n$ over $\mathbb{R}$ is attained at a point of the interval $\left(0,\frac{\pi}{n}\right)$. Due to the approximation $\sin(x)\leq x e^{-x^2/6}$ over the interval $(0,\pi)$ we have
$$ \sup g_n(x) \leq \sup n! x^n \exp\left(-\frac{n(n+1)(2n+1)}{36}x^2\right)= n!\left(\frac{18}{e(n+1)(2n+1)}\right)^{n/2}$$
and 
$$ \sup g_n(x) \leq \frac{6}{5}\sqrt{n}\,e^{-\frac{2}{5}n}. $$
It follows that on the interval $\left[-\frac{n}{5},\frac{n}{5}\right]$ the absolute value of $f_n(x)$ is bounded by $\frac{6}{5}e^{-n/5}$.
This bound implies U.C. towards zero on any compact subset of $\mathbb{R}$.
A: 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.
