Can this limit be solved with Riemann sum? 
Can this limit be solved with Riemann sum?
  $$ \lim _{n\to \infty }\left[\lim _{x\to 0}\left(\cos x\cdot\cos2x\cdots\cos nx\right)^{\frac{1}{n^3x^2}}\right] $$

What I've tried is to solve it with the Riemann sum but I am getting stuck somewhere , and I am not seeing where .
In my textbook I have the following options as answers :
$a. e^3$
$b. e^{-2}$
$c. e^{\frac{1}{3}}$
$d. e^{\frac{1}{2}}$
$ e. e^{\frac{-1}{6}}$

$$L=\lim_{n\rightarrow \infty} \lim_{x\rightarrow 0} (\cos x \cos 2x \cos 3x...\cos nx)^{\frac{1}{n^3x^2}}$$
As per @marty cohen, let us  use $-y+y^2/2 \le \ln(1-y)\le -y~$  and  $(1-t^2/2) \le \cos t \le (1-t^2/2+t^4/24)~$  when $~t~$ and $~y~$ are very small. We get
 $$\ln L =\lim_{n\rightarrow \infty} \lim_{x\rightarrow 0} \frac{1}{n^3 x^2} \sum_{r=1}^{n} (\ln \cos rx)=\lim_{n\rightarrow \infty} \lim_{x\rightarrow}\frac{1}{n^3 x^2}  \sum_{r=1}^n \ln(1-r^2x^2/2).$$ $$\Rightarrow \ln L = \lim_{n\rightarrow \infty} \lim_{x\rightarrow} \frac{1}{n^3 x^2} \sum_{r=1}^n (-r^2x^2/2)=\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{r=1}^n -r^2/(2n^2)=
\int_{0}^{1}(-z^2/2) dz=\frac{-1}{6}.$$ So $L=e^{-1/6}.$
