Question

$\underset{x\to 0}{\lim }\frac{\cos 2x-1}{\cos x-1}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{\cos \left(2x\right)-1}{\cos x-1}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{1-2{\sin }^{2}x-1}{1-2{\sin }^2\left({\displaystyle \frac{x}{2}}\right)-1}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\sin }^{2}x}{{\sin }^2\left({\displaystyle \frac{x}{2}}\right)}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\sin }^{2}x}{x^2}\times \frac{x^2}{{\displaystyle \frac{{\sin }^2\left({\displaystyle \frac{x}{2}}\right)}{{\displaystyle \frac{x^2}{4}}}\times \frac{x^2}{4}}}\right] \\ =\underset{x\to 0}{\lim }\left[{\left(\frac{\sin x}{x}\right)}^2\times \frac{1}{{\left({\displaystyle \frac{\sin \left({\displaystyle \frac{x}{2}}\right)}{{\displaystyle \frac{x}{2}}}}\right)}^2}\times 4\right] \\ =4 \end{gathered}$$