Let Me Give You More Trouble

A slightly different solution...

When $x$ goes to 0, $cos(kx)$ goes to $1 - \frac{(kx)^2}{2}$ . It is important to write the $x^2$ term, since it appears on the denominator. So:

$\large \displaystyle = \lim_{x\rightarrow 0} \frac{1 - \prod_{k=1}^n (1 - \frac{(kx)^2}{2})} {x^2}$

$\large \displaystyle = \lim_{x\rightarrow 0} \frac{1 - (1 - \sum_{k=1}^n \frac{(kx) ^2} {2} + O(x^4)) } {x^2}$

Since the terms $x^k, k > 2$ goes to $0$ faster than $x^2$ for $x \rightarrow 0$ , we're looking only for the $x^2$ terms in the product. So:

$\large \displaystyle = \lim_{x\rightarrow 0} \frac{1 - (1 - \sum_{k=1}^n \frac{(kx) ^2} {2}) } {x^2}$

The $x^2$ goes outside the sum and cancels out with the denominator:

$\large \displaystyle = \frac12 \sum_{k=1}^n k^2$

$\large \displaystyle = \frac{n(n+1)(2n+1)}{12}$

For $n = 100$ this is equal to $\color{#3D99F6} \boxed{ \large \displaystyle 169175}$

Relevant wiki: L'Hopital's Rule - Basic

Let $f(x) = \displaystyle\prod_{k=1}^{n}\cos{kx}$

$\begin{aligned} \log{(f(x))} & = \displaystyle\sum_{k=1}^{n} \log{(\cos{kx})} \hspace{5mm} \small\text{ differentiate w.r.t x on both sides } \\ \dfrac{1}{f(x)} \cdot f^{'}(x) & = \displaystyle\sum_{k=1}^{n} \dfrac{1}{\cos(kx)} \cdot -\left( \sin{(kx) } \right) \cdot k \\ f^{'}(x) & = -f(x) \displaystyle\sum_{k=1}^{n} k \tan{(kx)} \\ f^{''}(x) & =-f^{'}(x) \displaystyle\sum_{k=1}^{n} k\tan{(kx)}-f(x) \displaystyle\sum_{k=1}^{n} k^2 \sec^{2}{(kx)} \end{aligned}$

Now , $L = \displaystyle \lim_{x \to 0 } \dfrac{1-f(x)}{x^2} = \lim_{x \to 0 } \dfrac{-f^{'}(x)}{2x} = \lim_{x \to 0 } \dfrac{-f^{''}(x)}{2} = \dfrac{-f^{''}(0)}{2}$

$f^{''}(0) = 0 - 1 \cdot \displaystyle\sum_{k=1}^{n} k^2 \cdot 1 = -\dfrac{n(n+1)(2n+1)}{6}$

$\boxed{L = \dfrac{n(n+1)(2n+1)}{12}}$

Relevant wiki: L'Hopital's Rule - Basic

$\begin{aligned} L_n & = \lim_{x \to 0} \frac {1-\prod_{k=1}^n \cos kx}{x^2} & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies.} \\ & = \lim_{x \to 0} \frac {\prod_{k=1}^n \cos kx\sum_{j=1}^n j \tan jx}{2x} & \small \color{#3D99F6} \text{Differentiate up and down; 0/0 case again.} \\ & = \lim_{x \to 0} \frac {- \prod_{k=1}^n \cos kx \left(\sum_{j=1}^n j \tan jx\right)^2 + \prod_{k=1}^n \cos kx\sum_{j=1}^n j^2 \sec^2 jx }2 & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \text{ again.} \\ & = \frac {\sum_{j=1}^n j^2}2 \\ & = \frac {n(n+1)(2n+1)}{12} \end{aligned}$

Therefore, for $n=100$ , $L_{100} = \dfrac {100(101)(201)}{12} = \boxed{169175}$ .

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Note:

$\begin{aligned} f(x) & = \prod_{k=1}^n \cos kx = \cos(x)\cos(2x) \cos(3x) \cdots \cos(nx) \\ \frac {df(x)}{dx} & = \sin ({\color{#D61F06}x}) \prod_{\color{#D61F06} k\ne 1}^n \cos kx + 2 \sin ({\color{#D61F06}2x}) \prod_{\color{#D61F06} k\ne 2}^n \cos kx + \cdots + n \sin ({\color{#D61F06}n x}) \prod_{\color{#D61F06} k\ne n}^n \cos kx \\ & = \frac {\sin (x)}{\cos (x)} \prod_{\color{#3D99F6} k = 1}^n \cos kx + \frac {2\sin (2x)}{\cos (2x)} \prod_{\color{#3D99F6} k = 1}^n \cos kx + \cdots + \frac {n\sin (nx)}{\cos (nx)} \prod_{\color{#3D99F6} k = 1}^n \cos kx \\ & = \prod_{k = 1}^n \cos kx \sum_{j=1}^n j\tan jk \end{aligned}$