Evaluate: $\lim\limits_{x \to 0} \; \dfrac{\cos 2x -1}{\cos x -1}$
$\begin{aligned} \lim\limits_{x \to 0} \; \dfrac{\cos 2x -1}{\cos x -1} & = \lim\limits_{x \to 0} \; \dfrac{-2\sin^2 x}{-2 \sin^2 \left(\dfrac{x}{2}\right)} \;\; \left[\text{Note: }\cos 2 \theta -1 = -2 \sin^2 \theta\right] \\ & \\ & = \lim\limits_{x \to 0} \dfrac{\sin^2 x}{\sin^2 \left(\dfrac{x}{2}\right)} \\ & \\ & = \lim\limits_{x \to 0} \dfrac{\left(\dfrac{\sin x}{x}\right)^2}{\left[\dfrac{\sin \left(\dfrac{x}{2}\right)}{\dfrac{x}{2}}\right]^2 \times \dfrac{1}{4}} \\ & \\ & = 4 \times \dfrac{1^2}{1^2} = 4 \end{aligned}$