Question

$\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\sqrt{2}-\cos x-\sin x}{{\left(4x-\mathrm{\pi }\right)}^2}$

Answer 1

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\sqrt{2}-\cos x-\sin x}{{\left(4x-\mathrm{\pi }\right)}^2} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{2}-\left\{\cos \left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)+\sin \left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)\right\}}{{\left(4\left({\displaystyle \frac{\mathrm{\pi }}{4}}+h\right)-\mathrm{\pi }\right)}^2} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{2}-\left\{\cos \frac{\mathrm{\pi }}{4}\cos h-\sin \frac{\mathrm{\pi }}{4}\sin h+\sin \frac{\mathrm{\pi }}{4}\cos h+\cos \frac{\mathrm{\pi }}{4}\sin h\right\}}{{\left(4h\right)}^2} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{2}-\sqrt{2}\cos h}{{\left(4h\right)}^2} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{2}\left(1-\cos h\right)}{16h^2} \\ =\underset{h\to 0}{\lim }\frac{2\sqrt{2}{\sin }^2{\displaystyle \frac{h}{2}}}{64\times {\displaystyle \frac{h^2}{4}}} \\ =\frac{\sqrt{2}}{32}\times {\left(1\right)}^2 \\ =\frac{1}{16\sqrt{2}} \end{gathered}$$