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Question Number 176721 by youssefelaour last updated on 25/Sep/22

Answered by MJS_new last updated on 25/Sep/22

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

Answered by cortano1 last updated on 26/Sep/22

$$\underset{x\to 0}{\lim }\frac{2\cos (\mathrm{x}+\frac{\pi }{4})-\sqrt{2}}{2\sin (\mathrm{x}+\frac{\pi }{4})-\sqrt{2}}$$$$=\underset{x\to 0}{\lim }\frac{2\{\frac{1}{2}\sqrt{2}\cos \mathrm{x}-\frac{1}{2}\sqrt{2}\sin \mathrm{x}\}-\sqrt{2}}{2(\frac{1}{2}\sqrt{2}\sin \mathrm{x}+\frac{1}{2}\sqrt{2}\cos \mathrm{x})-\sqrt{2}}$$$$=\underset{x\to 0}{\lim }\frac{\sqrt{2}(\cos \mathrm{x}-1)-2\sqrt{2}\sin \frac{1}{2}\mathrm{x}\cos \frac{1}{2}\mathrm{x}}{2\sqrt{2}\sin \frac{1}{2}\mathrm{x}\cos \frac{1}{2}\mathrm{x}+\sqrt{2}(\cos \mathrm{x}-1)}$$$$=\underset{x\to 0}{\lim }\frac{-2\sqrt{2}\sin \frac{1}{2}\mathrm{x}(\sin \frac{1}{2}\mathrm{x}+\cos \frac{1}{2}\mathrm{x})}{2\sqrt{2}\sin \frac{1}{2}\mathrm{x}(\cos \frac{1}{2}\mathrm{x}-\sin \frac{1}{2}\mathrm{x})}$$$$=\underset{x\to 0}{\lim }\frac{\sin \frac{1}{2}\mathrm{x}+\cos \frac{1}{2}\mathrm{x}}{\sin \frac{1}{2}\mathrm{x}-\cos \frac{1}{2}\mathrm{x}}=-1$$

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