Ta có
$I = \underset{x \to 0}{\lim} \dfrac{\cos x - \cos(3x)}{\sin^2x}$
$= -2 \underset{x \to 0}{\lim} \dfrac{\sin(2x) \sin (-x)}{\sin^2x}$
$= 2 \underset{x \to 0}{\lim} \dfrac{2\sin x \cos x}{\sin x}$
$= 2 \underset{x \to 0}{\lim} 2 \cos x = 2.2 = 4$