To demonstrate a method for computing the limit itself, let's pick a small value of n. If n = 3, then our limit is
[tex]\displaystyle \lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{3} \sqrt[k]{\cos(kx)} }{ {x}^{2} }[/tex]
Let a = 1 and b the cosine product, and write them as
[tex]\dfrac{a - b}{x^2}[/tex]
with
[tex]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}[/tex]
Now we use the identity
[tex]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)[/tex]
to rationalize the numerator. This gives
[tex]\displaystyle \frac{a^6-b^6}{x^2 \left(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5\right)}[/tex]
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
[tex]\displaystyle \frac16 \lim_{x\to0} \frac{1-\cos^3(2x)\cos^2(3x)}{x^2}[/tex]
For the remaining limit, use the Taylor expansion for cos(x) :
[tex]\cos(x) = 1 - \dfrac{x^2}2 + \mathcal{O}(x^4)[/tex]
where [tex]\mathcal{O}(x^4)[/tex] 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
[tex]\displaystyle \cos^3(2x) \cos^2(3x) = \left(1 - 2x^2\right)^3 \left(1 - \frac{9x^2}2\right)^2[/tex]
[tex]\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)[/tex]
[tex]\displaystyle \cos^3(2x) \cos^2(3x) = 1 - 15x^2 + \mathcal{O}(x^4)[/tex]
so in our limit, the constant terms cancel, and the asymptotic terms go to 0, and we end up with
[tex]\displaystyle \frac16 \lim_{x\to0} \frac{15x^2}{x^2} = \frac{15}6 = \frac52[/tex]
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.
[tex]\displaystyle \rm\lim_{x \to 0 } \frac{1 - \prod \limits_{k = 2}^{n} \sqrt[k]{cos(kx)} }{ {x}^{2} } = 10[/tex]
[tex] \displaystyle\rm Let \: y = \prod_{k = 2}^n \sqrt[k]{cos (kx)} [/tex]
[tex] \to \rm logy = log \prod \limits _{k = 1}^{n} \sqrt[k]{cos(kx)} [/tex]
[tex] { \to\rm logy = \sum \limits_{k = 2}^{n} log( \sqrt[k]{ \cos(kx) } ) = \sum \limits_{k = 2}^{n} \frac{1}{k} \log( \cos(kx) ) }[/tex]
[tex] \rm differentiate \: wrtx[/tex]
[tex] {\displaystyle \to \rm \frac{1}{y} \: dy = \sum_{k = 2}^{ n } \bigg\{ \frac{1}{ \cancel{k}} \cdot\frac{1}{ \cos(kx) } \cdot - \cancel{k} \sin(kx) \bigg \}}[/tex]
[tex] \displaystyle \to \rm dy = - y \sum_{k = 2}^{n} \tan(kx) [/tex]
[tex] \rm{Now}[/tex]
[tex] \displaystyle \rm \lim_{x \to0} \frac{1 - y}{ {x}^{2} } = \frac{ - dy}{2x} \: \: \: \: [using - Hopitals \: rule ][/tex]
[tex] \displaystyle \rm = \lim_{x \to0} \frac{y}{ {2} } \sum_{k = 2}^{n} \frac{ \tan(kx) }{kx} \times k[/tex]
[tex]\displaystyle \rm = \lim_{x \to0} \frac{y}{ {2} } \sum_{k = 2}^{n} k \: \: \: \: \: \bigg [note \: \displaystyle \rm\lim_{x \to0} \: y = 1 \bigg][/tex]
[tex] \rm= \dfrac{1}{2} \bigg [ \dfrac{n(n + 1)}{2} - 1\bigg] = 10[/tex]
[tex] \rm \to n(n + 1) = 2(20 + 1) = 42 = 6 \times 7 = 6(6 + 1)[/tex]
[tex] \rm \boxed{ \rm{n = 6}}[/tex]
