Question

$\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{1-\sin 2x}{1+\cos 4x}$

Answer 1

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{1-\sin 2x}{1+\cos 4x} \\ =\underset{h\to 0}{\lim }\frac{1-\sin 2\left({\displaystyle \frac{\mathrm{\pi }}{4}}-h\right)}{1+\cos 4\left({\displaystyle \frac{\mathrm{\pi }}{4}}-h\right)} \\ =\underset{h\to 0}{\lim }\frac{1-\cos 2h}{1-\cos 4h} \\ =\underset{h\to 0}{\lim }\frac{2{\sin }^{2}h}{2{\sin }^{2}2h} \\ \Rightarrow \underset{h\to 0}{\lim }\frac{{\displaystyle \frac{{\sin }^{2}h}{h^2}}}{{\displaystyle \frac{4{\sin }^{2}2h}{4h^2}}} \\ \Rightarrow \frac{1}{4} \end{gathered}$$