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Relevant wiki: Maclaurin Series

The problem can be solved by using Maclaurin series of $\sin x$ twice as follows.

$\begin{aligned} L & = \lim_{x \to 0} \frac {2\sin (\sin 2x)-\sin (\sin 4x)}{x^3} \\ & = \lim_{x \to 0} \frac 1{x^3} \left({\color{#3D99F6}2\sin \left(2x - \frac {2^3x^3}{3!} + \frac {2^5x^5}{5!} - \cdots \right)} - \color{#D61F06} \sin \left(4x - \frac {4^3x^3}{3!} + \frac {4^5x^5}{5!} - \cdots \right) \right) \\ & = \lim_{x \to 0} \frac 1{x^3} \left({\color{#3D99F6}\left(4x - \frac {16x^3}{3!} + O(x^5) - \frac {16x^3}{3!} + O(x^5) \right)} - \color{#D61F06} \left(4x - \frac {64x^3}{3!} + O(x^5) - \frac {64x^3}{3!} + O(x^5) \right) \right) \\ & = \lim_{x \to 0} \frac 1{x^3} \left(\frac {96x^3}{3!} + O(x^5) \right) \\ & = \lim_{x \to 0} \left(16 + O(x^2) \right) \\ & = \boxed {16} \end{aligned}$

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