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Question Number 132240 by bemath last updated on 12/Feb/21

$$\underset{x\to 0}{\lim }\frac{\tan (2\mathrm{x}+\frac{\pi }{4})-2\tan (\mathrm{x}+\frac{\pi }{4})+\tan \frac{\pi }{4}}{\sin (2\mathrm{x}+\frac{\pi }{4})-2\sin (\mathrm{x}+\frac{\pi }{4})+\sin \frac{\pi }{4}}?$$$$$$

Answered by EDWIN88 last updated on 13/Feb/21

$$\underset{x\to 0}{\lim }\frac{2\sec {}^2(2\mathrm{x}+\frac{\pi }{4})-2\sec {}^2(\mathrm{x}+\frac{\pi }{4})}{2\cos (2\mathrm{x}+\frac{\pi }{4})-2\cos (\mathrm{x}+\frac{\pi }{4})}$$$$=\underset{x\to 0}{\lim }\frac{\cos {}^2(\mathrm{x}+\frac{\pi }{4})-\cos {}^2(2\mathrm{x}+\frac{\pi }{4})}{{[\cos (2\mathrm{x}+\frac{\pi }{4}).\cos (\mathrm{x}+\frac{\pi }{4})]}^2[\cos (2\mathrm{x}+\frac{\pi }{4})-\cos (\mathrm{x}+\frac{\pi }{4})]}$$$$=\underset{x\to 0}{\lim }\frac{-[\cos (\mathrm{x}+\frac{\pi }{4})+\cos (2\mathrm{x}+\frac{\pi }{4})]}{{[\cos (2\mathrm{x}+\frac{\pi }{4}).\cos (\mathrm{x}+\frac{\pi }{4})]}^2}$$$$=-\frac{\sqrt{2}}{{\left(\frac{1}{2}\right)}^2}=-4\sqrt{2}$$

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