Answer
$\dfrac{1}{2}$
Work Step by Step
L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$
Here, $\lim\limits_{x \to 0} f(0)=\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to 0} \dfrac{3 \cos 3x-3+2x}{(\cos 2x)\sin 2x+(2\cos 2x) \sin x}=\dfrac{0}{0}$
This shows an indeterminate form of the limit, so we need to use L'Hospital's rule.
$\lim\limits_{x \to 0} \dfrac{2-9 \sin 3x}{-5\sin 2x\sin x+4 \cos 2x \cos x}=\dfrac{2}{0+4}=\dfrac{1}{2}$
