<s>[INST] Points of discontinuity of $\left[\frac{f(x)}3\right], f(x)=\lim_{n\to\infty}\ln(\sqrt{e^{\cos x}\sqrt{e^{3\cos x}...\sqrt{e^{(2n+1)\cos x}}}})$ 
Let $f(x)=\lim_{n\to\infty}\ln\left(\sqrt{e^{\cos x}\sqrt{e^{3\cos x}\sqrt{e^{5\cos x}...\sqrt{e^{(2n+1)\cos x}}}}}\right)$ and if $g(x)=\left[\frac{f(x)}3\right]$, then the number of points in $[0,2\pi]$ where $g(x)$ is discontinuous, is/are (where $[*]$ represents greatest integer function)...

$\cos x$ varies from $-1$ to $1$.
If $\cos x=-1$ then as $n\to\infty$, $e^{(2n+1)\cos x}=0$
If $\cos x=0$ then as $n\to\infty$, $e^{(2n+1)\cos x}=1$
If $\cos x=\infty$ then as $n\to\infty$, $e^{(2n+1)\cos x}=\infty$
Also, since $\cos x$ is changing these values in continuous manner, can we use that to comment about the continuity of $f(x)$?
But before that, I would need to solve the limit. Not able to do so. Can you give any pointers? Thanks.
EDIT: $f(x)=\lim_{n\to\infty}\ln\left(e^{\left(\frac12+\frac3{2^2}+\frac5{2^3}+...+\frac{(2n+1)}{2^{n+1}}\right)\cos x}\right)$
Let $$S=\frac12+\frac3{2^2}+\frac5{2^3}+...+\frac{(2n+1)}{2^{n+1}}\\\frac S2=\frac1{2^2}+\frac3{2^3}+\frac5{2^4}+...+\frac{(2n+1)}{2^{n+2}}$$
Subtracting, $$\frac S2=\frac12+\frac12+...\text{(n+1) times}-\frac{2n+1}{2^{n+2}}=\frac {n+1}2-\frac{2n+1}{2^{n+2}}\\\implies S=n+1-\frac{2n+1}{2^{n+1}}$$
I think my $S$ is coming out to be $\infty$. Can you point out the mistake? Thanks.
EDIT$2$: After reading the comments, I reworked and found $S=3$.
Thus, $f(x)=3\cos x\implies\left[\frac{f(x)}3\right]=[\cos x]$
Thus, there are $4$ points of discontinuity $\{0,\frac\pi2,\frac{3\pi}2,2\pi\}$.
 [/INST] Hint:
Let $r = e^{\cos(x)}$. Then verify by induction or otherwise that
$$ \sqrt{r \sqrt{r^3 \sqrt{r^5 \ldots \sqrt{r ^{2n+1}  }}} } =  r^{\frac12+ \frac{3}{2^2} + \ldots+ \frac{2n+1}{2^{n+1}} }$$
Now evaluate $S(n) = \sum_{k=0}^{n}\frac{2k+1}{2^{k+1}}$ using this and geometric series.
Then evaluate $f(x) = \lim_{n\to \infty} \ln( e^{S(n)\cos(x)}) = \lim_{n\to \infty} S(n)\cos (x)$ then proceed to find the answer.
Update
I think there is a problem in calculating $S(n)$. To evaluate it
use the formula given here, then we see that
$$ \sum_{k=0}^{n}\frac{k}{2^k} = \frac{2^{n + 1} - n - 2}{2^{n}}$$
So $$S(n) = \frac{{3 \cdot 2^{n + 1} - 2 \, n - 5}} {2^{n + 1}}$$
Now can you find the $\lim_{n\to \infty}S(n)$?
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