To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is
![\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%200%20%7D%20%5Cfrac%7B1%20-%20%5Cprod%20%5Climits_%7Bk%20%3D%202%7D%5E%7B3%7D%20%5Csqrt%5Bk%5D%7B%5Ccos%28kx%29%7D%20%7D%7B%20%7Bx%7D%5E%7B2%7D%20%7D)
Let a = 1 and b the cosine product, and write them as
![\dfrac{a - b}{x^2}](https://tex.z-dn.net/?f=%5Cdfrac%7Ba%20-%20b%7D%7Bx%5E2%7D)
with
![b = \sqrt{\cos(2x)} \sqrt[3]{\cos(3x)} = \sqrt[6]{\cos^3(2x)} \sqrt[6]{\cos^2(3x)} = \left(\cos^3(2x) \cos^2(3x)\right)^{\frac16}](https://tex.z-dn.net/?f=b%20%3D%20%5Csqrt%7B%5Ccos%282x%29%7D%20%5Csqrt%5B3%5D%7B%5Ccos%283x%29%7D%20%3D%20%5Csqrt%5B6%5D%7B%5Ccos%5E3%282x%29%7D%20%5Csqrt%5B6%5D%7B%5Ccos%5E2%283x%29%7D%20%3D%20%5Cleft%28%5Ccos%5E3%282x%29%20%5Ccos%5E2%283x%29%5Cright%29%5E%7B%5Cfrac16%7D)
Now we use the identity
![a^n-b^n = (a-b)\left(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots a^2b^{n-3}+ab^{n-2}+b^{n-1}\right)](https://tex.z-dn.net/?f=a%5En-b%5En%20%3D%20%28a-b%29%5Cleft%28a%5E%7Bn-1%7D%2Ba%5E%7Bn-2%7Db%2Ba%5E%7Bn-3%7Db%5E2%2B%5Ccdots%20a%5E2b%5E%7Bn-3%7D%2Bab%5E%7Bn-2%7D%2Bb%5E%7Bn-1%7D%5Cright%29)
to rationalize the numerator. This gives
![\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Ba%5E6-b%5E6%7D%7Bx%5E2%20%5Cleft%28a%5E5%2Ba%5E4b%2Ba%5E3b%5E2%2Ba%5E2b%5E3%2Bab%5E4%2Bb%5E5%5Cright%29%7D)
As x approaches 0, both a and b approach 1, so the polynomial in a and b in the denominator approaches 6, and our original limit reduces to
![\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac16%20%5Clim_%7Bx%5Cto0%7D%20%5Cfrac%7B1-%5Ccos%5E3%282x%29%5Ccos%5E2%283x%29%7D%7Bx%5E2%7D)
For the remaining limit, use the Taylor expansion for cos(x) :
![\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)](https://tex.z-dn.net/?f=%5Ccos%28x%29%20%3D%201%20-%20%5Cdfrac%7Bx%5E2%7D2%20%2B%20%5Cmathcal%7BO%7D%28x%5E4%29)
where
essentially means that all the other terms in the expansion grow as quickly as or faster than x⁴; in other words, the expansion behaves asymptotically like x⁴. As x approaches 0, all these terms go to 0 as well.
Then
![\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ccos%5E3%282x%29%20%5Ccos%5E2%283x%29%20%3D%20%5Cleft%281%20-%202x%5E2%5Cright%29%5E3%20%5Cleft%281%20-%20%5Cfrac%7B9x%5E2%7D2%5Cright%29%5E2)
![\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 6x^2 + 12x^4 - 8x^6\right) \left(1 - 9x^2 + \frac{81x^4}4\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ccos%5E3%282x%29%20%5Ccos%5E2%283x%29%20%3D%20%5Cleft%281%20-%206x%5E2%20%2B%2012x%5E4%20-%208x%5E6%5Cright%29%20%5Cleft%281%20-%209x%5E2%20%2B%20%5Cfrac%7B81x%5E4%7D4%5Cright%29)
![\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ccos%5E3%282x%29%20%5Ccos%5E2%283x%29%20%3D%201%20-%2015x%5E2%20%2B%20%5Cmathcal%7BO%7D%28x%5E4%29)
so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with
![\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac16%20%5Clim_%7Bx%5Cto0%7D%20%5Cfrac%7B15x%5E2%7D%7Bx%5E2%7D%20%3D%20%5Cfrac%7B15%7D6%20%3D%20%5Cfrac52)
Unfortunately, this doesn't agree with the limit we want, so n ≠ 3. But you can try applying this method for larger n, or computing a more general result.
Edit: some scratch work suggests the limit is 10 for n = 6.