$\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{lim} \frac{8\sqrt{2}-(\cos \mathrm{x}+\sin \mathrm{x}{)}^7}{\sqrt{2}-\sqrt{2}\sin 2\mathrm{x}}$ is equal to

$$\begin{gathered} \underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{lim} \frac{8\sqrt{2}-(\cos \mathrm{x}+\sin \mathrm{x}{)}^7}{\sqrt{2}-\sqrt{2}\sin 2\mathrm{x}} \\ =\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{lim} \frac{0-7(\cos \mathrm{x}+\sin \mathrm{x}{)}^6\left(-\sin x+\cos x\right)}{0-2\sqrt{2}\cos 2\mathrm{x}} \\ =\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{lim} \frac{-7(\cos \mathrm{x}+\sin \mathrm{x}{)}^6\left(\cos x-\sin x\right)}{-2\sqrt{2}\left({\cos }^{2}x-{\sin }^{2}x\right)} \\ =\underset{\mathrm{x}\to \frac{\mathrm{\pi }}{4}}{lim} \frac{-7(\cos \mathrm{x}+\sin \mathrm{x}{)}^6}{-2\sqrt{2}\left(\cos x+\sin x\right)}=\frac{7{\left({\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{1}{\sqrt{2}}}\right)}^5}{2\sqrt{2}}=\frac{7.4\sqrt{2}}{2\sqrt{2}}=14 \end{gathered}$$