Question Number 132240 by bemath last updated on 12/Feb/21
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{tan}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{2tan}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{tan}\:\frac{\pi}{\mathrm{4}}}{\mathrm{sin}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{2sin}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{sin}\:\frac{\pi}{\mathrm{4}}}\:? \\ $$$$ \\ $$
Answered by EDWIN88 last updated on 13/Feb/21
![lim_(x→0) ((2sec^2 (2x+(π/4))−2sec^2 (x+(π/4)))/(2cos (2x+(π/4))−2cos (x+(π/4)))) = lim_(x→0) ((cos^2 (x+(π/4))−cos^2 (2x+(π/4)))/([ cos (2x+(π/4)).cos (x+(π/4))]^2 [cos (2x+(π/4))−cos (x+(π/4))])) = lim_(x→0) ((−[ cos (x+(π/4))+cos (2x+(π/4)) ])/([ cos (2x+(π/4)).cos (x+(π/4))]^2 )) = −((√2)/(((1/2))^2 )) = −4(√2)](https://www.tinkutara.com/question/Q132284.png)
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sec}\:^{\mathrm{2}} \left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{2sec}\:^{\mathrm{2}} \left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{2cos}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{2cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)}{\left[\:\mathrm{cos}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right).\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)\right]^{\mathrm{2}} \left[\mathrm{cos}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)\right]} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\left[\:\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{cos}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)\:\right]}{\left[\:\mathrm{cos}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right).\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)\right]^{\mathrm{2}} } \\ $$$$=\:−\frac{\sqrt{\mathrm{2}}}{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\:=\:−\mathrm{4}\sqrt{\mathrm{2}} \\ $$