\begin{aligned} \lim_{x\to 0} \ \left( \frac{\sin 4x}{x^2 \tan 2x} - \frac{2}{x^2} \right) &= \lim_{x\to 0} \ \left( \frac{\sin 4x}{x^2 \tan 2x} - \frac{2 \tan 2x}{x^2 \tan 2x} \right) = \lim_{x\to 0} \ \frac{\sin 4x - 2\tan 2x}{x^2 \tan 2x} \\[8pt] &= \lim_{x\to 0} \ \frac{\sin 4x - 2 \cdot \frac{\sin 2x}{\cos 2x}}{x^2 \tan 2x} = \lim_{x\to 0} \ \frac{\sin (2 \cdot 2x) \ \cos 2x - 2\sin 2x}{x^2 \tan 2x \ \cos 2x} \\[8pt] &= \lim_{x\to 0} \ \frac{(2 \sin 2x \cos 2x) \cos 2x - 2\sin 2x}{x^2 \ \frac{\sin 2x}{\cos 2x} \cdot \cos 2x} = \lim_{x\to 0} \ \frac{2 \sin 2x (\cos^2 2x - 1)}{x^2 \sin 2x} \\[8pt] &= \lim_{x\to 0} \ \frac{2 \sin 2x \ (-\sin^2 2x)}{x^2 \sin 2x} = \lim_{x\to 0} \ \frac{-2 \sin^2 2x }{x^2} \\[8pt] &= -2 \cdot \lim_{x\to 0} \ \frac{\sin 2x }{x} \cdot \lim_{x\to 0} \ \frac{\sin 2x }{x} \\[8pt] &= -2 \cdot 2 \cdot 2 \\[8pt] &= -8 \end{aligned}