Question

$\underset{x\to 0}{\lim }\frac{1-\cos 2x}{\cos 2x-\cos 8x}$

Answer 1

$\underset{x\to 0}{\lim }\left[\frac{1-\cos 2x}{\cos 2x-\cos 8x}\right]$

$$\begin{gathered} =\underset{x\to 0}{\lim }\left[\frac{2{\sin }^{2}x}{2\sin \left({\displaystyle \frac{2x+8x}{2}}\right)\sin \left({\displaystyle \frac{8x-2x}{2}}\right)}\right]\left[\because \mathrm{cosC}-\mathrm{cosD}=2\sin \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right)\sin \left(\frac{D-C}{2}\right)\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2\sin x\times \sin x}{2\sin 5x\times \sin 3x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\sin x\times \sin x}{\sin 5x\times \sin 3x}\right]\left[\because \sin \left(-\theta \right)=-\sin \theta \right] \\ =\underset{x\to 0}{\lim }\left[\frac{{\displaystyle \frac{\sin x}{x}}\times x\times {\displaystyle \frac{\sin x}{x}}\times x}{{\displaystyle \frac{\sin 5x}{5x}}\times 5x\times {\displaystyle \frac{\sin 3x\times 3x}{3x}}}\right] \\ =\frac{1}{15} \end{gathered}$$