Evaluate $\lim\limits_{x \to 0} \; \dfrac{\cos x -1}{\cos 2x -1}$
$\begin{aligned} & \lim\limits_{x \to 0} \; \dfrac{\cos x -1}{\cos 2x -1} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\cos x -1\right)}{\dfrac{d}{dx}\left(\cos 2x -1\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{-\sin x}{-2\sin 2x} & \left(\dfrac{0}{0} \text{ form}\right) \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\dfrac{d}{dx}\left(\sin x\right)}{\dfrac{d}{dx}\left(2 \sin 2x\right)} & \left[\text{L'Hospital's rule}\right] \\ & & \\ & = \lim\limits_{x \to 0} \; \dfrac{\cos x}{4 \cos 2x} & \\ & & \\ & = \dfrac{1}{4} & \end{aligned}$