Tomamos como el límite
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(6 x \right)} - 1}{x^{2}}\right)$$
Usamos la fórmula trigonométrica
sin(a)^2 = (1 - cos(2*a))/2
cambiamos
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(6 x \right)} - 1}{x^{2}}\right) = \lim_{x \to 0^+}\left(- \frac{2 \sin^{2}{\left(3 x \right)}}{x^{2}}\right)$$
=
$$- 2 \left(\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x}\right)\right)^{2}$$
Sustituimos
$$u = 3 x$$
entonces
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x}\right) = \lim_{u \to 0^+}\left(\frac{3 \sin{\left(u \right)}}{u}\right)$$
=
$$3 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
El límite
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
hay el primer límite, es igual a 1.
entonces
$$- 2 \left(\lim_{x \to 0^+}\left(\frac{\sin{\left(3 x \right)}}{x}\right)\right)^{2} = - 2 \left(3 \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)\right)^{2}$$
=
$$- 2 \cdot 3^{2}$$
=
$$-18$$
Entonces la respuesta definitiva es:
$$\lim_{x \to 0^+}\left(\frac{\cos{\left(6 x \right)} - 1}{x^{2}}\right) = -18$$