Answer is: option3

\( 2 \cos 2x \)

Solution:

To solve the given limit:

\[ \lim\limits_{h \to 0} \frac{\sin 2(x+h) - \sin 2x}{h} \]

This limit represents the definition of the derivative:

\[ \frac{d}{dx} \sin 2x = \lim\limits_{h \to 0} \frac{\sin 2(x+h) - \sin 2x}{h} \] Using the derivative formula: \[ \frac{d}{dx} \sin 2x = 2 \cos 2x \]

Thus,

\[ \lim\limits_{h \to 0} \frac{\sin 2(x+h) - \sin 2x}{h} = 2 \cos 2x \] Comparing with the answer choices, the correct option is: \[ \boxed{2 \cos 2x} \]