Question

$\underset{x\to 0}{\lim }\frac{1-\cos 4x}{x^2}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{1-\cos \left(4x\right)}{x^2}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{2{\sin }^{2}2x}{x^2}\right] \\ =\underset{x\to 0}{\lim }2\left[\frac{\sin 2x}{x}\times \frac{\sin 2x}{x}\right] \\ =\underset{x\to 0}{\lim }\left[2\times \frac{\sin 2x}{2x}\times \frac{\sin 2x}{2x}\times 4\right] \\ =2\times 1\times 1\times 4 \\ =8 \end{gathered}$$