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Question Number 143995 by liberty last updated on 20/Jun/21

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

Answered by mathmax by abdo last updated on 20/Jun/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\pi -4\mathrm{x}}{\sqrt{1-\sqrt{\sin 2\mathrm{x}}}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right){=}_{\frac{\pi }{4}-\mathrm{x}=\mathrm{t}}\frac{\pi -4(\frac{\pi }{4}-\mathrm{t})}{\sqrt{1-\sqrt{\sin (2(\frac{\pi }{4}-\mathrm{t})}}}$$$$=\frac{4\mathrm{t}}{\sqrt{1-\sqrt{\sin (\frac{\pi }{2}-2\mathrm{t})}}}=\frac{4\mathrm{t}}{\sqrt{1-\sqrt{\cos \left(2\mathrm{t}\right)}}}=\mathrm{g}\left(\mathrm{t}\right)(\mathrm{t}\to 0)$$$$\cos \left(2\mathrm{t}\right)\sim 1-\frac{{4\mathrm{t}}^2}{2}=1-{2\mathrm{t}}^2\Rightarrow \sqrt{\cos \left(2\mathrm{t}\right)}\sim \sqrt{1-{2\mathrm{t}}^2}\sim 1-\frac{1}{2}\left({2\mathrm{t}}^2\right)=1-{\mathrm{t}}^2$$$$-\sqrt{\cos \left(2\mathrm{t}\right)}\sim {\mathrm{t}}^2-1\Rightarrow 1-\sqrt{\cos \left(2\mathrm{t}\right)}\sim {\mathrm{t}}^2\Rightarrow \sqrt{1-\sqrt{\cos \left(2\mathrm{t}\right)}}\sim \mathrm{t}\Rightarrow$$$$\mathrm{g}\left(\mathrm{t}\right)\sim \frac{4\mathrm{t}}{\mathrm{t}}\Rightarrow {\lim }_{\mathrm{t}\to 0}\mathrm{g}\left(\mathrm{t}\right)=4$$

Commented by liberty last updated on 21/Jun/21

$$\mathrm{in}\mathrm{my}\mathrm{book}\mathrm{the}\mathrm{limit}\mathrm{doesnot}\mathrm{exist}$$

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