Squeeze it repeatedly

Let $\dfrac{1}{L_n} = \displaystyle \lim _{x \to 0} \frac{f^{n} (x)}{x^m}$ for non-negative integers $n$ .

For $\frac{1}{L_{n}}$ to be a non-zero finite number, $m$ must be equal to the least power of $x$ in the Taylor series of $f^{n} (x)$ . $L_{n}$ will be equal to the reciprocal of the coefficient of least power of $x$ in the Taylor series of $f^{n} (x)$ .

$1 - \cos x$ can be written as $2 \sin^{2} \frac{x}{2}$ . When $x \to 0$ , $\sin x = x$ , therefore $1 - \cos x = \frac{x^{2}}{2}$ . Using this formula, $f^{n} (x)$ can be recursively defined as $f^{n} (x) = \frac{(f^{n - 1} (x))^{2}}{2}$ for very small $x$ . The Taylor series also yields the same results.

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When $n = 0$ , $f^{0} (x) = \frac{x^{\color{#3D99F6}{1}}}{\color{#D61F06}{1}}$ . Least power of $x$ is $\color{#3D99F6}{1}$ , so $m =\color{#3D99F6}{1}$ . $L_{0} =$ reciprocal of coefficient of $x$ $= \color{#D61F06}{1}$

When $n = 1$ , $f^{1} (x) = \frac{x^{\color{#3D99F6}{2}}}{\color{#D61F06}{2!}} - \frac{x^4}{4!} + \ldots \implies m = \color{#3D99F6}{2}, L_{1} = \color{#D61F06}{2}$

When $n = 2$ , $f^{2} (x) = \frac{x^{\color{#3D99F6}{4}}}{\color{#D61F06}{8}} + \ldots \implies m = \color{#3D99F6}{4}, L_{2} = \color{#D61F06}{8}$

On solving the recurrence relation $L_{n} = 2 (L_{n-1} )^{2}$ and satisfying the initial conditions, we get the following result.

For general $n$ , $f^{n} (x) = \frac{x^{\color{#3D99F6}{2^{n}}}}{\color{#D61F06}{2 ^{2^{n} - 1}}} + \ldots \implies m =\color{#3D99F6}{2^{n}}, L_{n} = \color{#D61F06}{2^{2^{n} - 1}}$

Now we substitute $n = 10000$ .

$2 L_{10000} = 2^{2^{10000}}$

$\log_{2} (2L) = 2^{10000}$

$\log_{32} \log_{2} (2L) = \log_{32} 2^{10000} = 10000 \log _{32} 2 = \frac{10000}{5} = \boxed{2000}$ $_\square$