calculus - 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?

Answer

Hint: $\sin(x)$ is always between 1 and 0. But the denominator keeps getting larger as n goes to infinity.