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Question Number 140574 by john_santu last updated on 09/May/21

$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{1-\tan {}^{3}x}{\cos 2x}=?$$

Answered by bramlexs22 last updated on 09/May/21

$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{(1+\tan {}^2\mathrm{x})(1-\tan {}^3\mathrm{x})}{1-\tan {}^2\mathrm{x}}$$$$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{1+\tan {}^2\mathrm{x}}{1+\tan \mathrm{x}}.\underset{x\to \frac{\pi }{4}}{\lim }\frac{(1-\tan \mathrm{x})(\tan {}^2\mathrm{x}+\tan \mathrm{x}+1)}{1-\tan \mathrm{x}}$$$$=\frac{2}{2}\times \frac{3}{1}=3.$$

Answered by john_santu last updated on 09/May/21

$$\underset{x\to \frac{\pi }{4}}{\lim }\frac{\cos {}^{3}x-\sin {}^{3}x}{\cos {}^{3}x(\cos {}^{2}x-\sin {}^{2}x)}$$$$=\underset{x\to \frac{\pi }{4}}{\lim }\frac{\cos {}^{2}x+\cos x.\sin x+\sin {}^{2}x}{\cos {}^{3}x(\cos x+\sin x)}.\underset{x\to \frac{\pi }{4}}{\lim }\frac{\cos x-\sin x}{\cos x-\sin x}$$$$=\frac{\frac{3}{2}}{\frac{\sqrt{2}}{4}\left(\sqrt{2}\right)}=3.$$

Answered by EDWIN88 last updated on 09/May/21

$$\mathrm{x}=\frac{\pi }{4}+\mathrm{z}\Rightarrow \underset{\mathrm{z}\to 0}{\lim }\frac{1-\tan {}^3(\mathrm{z}+\frac{\pi }{4})}{\cos (2\mathrm{z}+\frac{\pi }{2})}$$$$=\underset{\mathrm{z}\to 0}{\lim }\frac{(\tan {}^2(\mathrm{z}+\frac{\pi }{4})+\tan (\mathrm{z}+\frac{\pi }{4})+1)(1-\tan (\mathrm{z}+\frac{\pi }{4}))}{-\sin 2\mathrm{z}}$$$$=3\times \underset{\mathrm{z}\to 0}{\lim }\frac{\cos (\mathrm{z}+\frac{\pi }{4})-\sin (\mathrm{z}+\frac{\pi }{4})}{-\cos (\mathrm{z}+\frac{\pi }{4})\sin 2\mathrm{z}}$$$$=-\frac{6}{\sqrt{2}}\times \underset{\mathrm{z}\to 0}{\lim }\frac{-\sin (\mathrm{z}+\frac{\pi }{4})-\cos (\mathrm{z}+\frac{\pi }{4})}{2\cos 2\mathrm{z}}$$$$=-\frac{6}{\sqrt{2}}\times \frac{-\sqrt{2}}{2}=3$$

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