Answer
$\dfrac{9}{2}$
Work Step by Step
Our aim is to evaluate the limit for $\lim\limits_{x \to 0} [\dfrac{1}{x^2}\dfrac{\cos (3x)}{x^2}]$
We can see that the numerator and denominator have a limit of $0$, so the limit shows the indeterminate form of type $\dfrac{0}{0}$. So, we will apply L-Hospital's rule which can be defined as: $\lim\limits_{x \to a} \dfrac{P(x) }{Q(x)}=\lim\limits_{x \to a}\dfrac{P'(x)}{Q'(x)}$
where, $a$ can be any real number, infinity or negative infinity.
$\lim\limits_{x \to 0} [\dfrac{1}{x^2}\dfrac{\cos (3x)}{x^2}]=\lim\limits_{x \to 0} \dfrac{3 \sin 3x}{2x}=\dfrac{0}{0}$
Again apply L-Hospital's rule to obtain:
So, $\lim\limits_{x \to 0} \dfrac{3 \sin 3x}{2x}=3 \times \lim\limits_{x \to 0} \dfrac{3 \cos 3x}{2}=\dfrac{9}{2}$