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Question Number 132079 by liberty last updated on 11/Feb/21
If L = lim_(x→π/4) ((tan^3 x−tan x)/(cos (x+(π/4))))  then (L/4) =?
$$\mathrm{If}\:\mathrm{L}\:=\:\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}−\mathrm{tan}\:\mathrm{x}}{\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\mathrm{then}\:\frac{\mathrm{L}}{\mathrm{4}}\:=? \\ $$
Answered by EDWIN88 last updated on 11/Feb/21
 L=lim_(x→π/4) ((tan x(tan x+1)(tan x−1))/(cos (x+(π/4))))  L= 1×2×lim_(x→π/4) (((sin x−cos x)/(cos x))).((1/(cos (x+(π/4)))))  L=−1×4×lim_(x→π/4) (((1/( (√2)))cos x−(1/( (√2)))sin x)/(cos (x+(π/4))))  L=−4 ×lim_(x→π/4) ((cos (x+(π/4)))/(cos (x+(π/4)))) = −4    then (L/4) = −1
$$\:\mathrm{L}=\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{tan}\:\mathrm{x}\left(\mathrm{tan}\:\mathrm{x}+\mathrm{1}\right)\left(\mathrm{tan}\:\mathrm{x}−\mathrm{1}\right)}{\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\mathrm{L}=\:\mathrm{1}×\mathrm{2}×\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\right).\left(\frac{\mathrm{1}}{\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)}\right) \\ $$$$\mathrm{L}=−\mathrm{1}×\mathrm{4}×\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{cos}\:\mathrm{x}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\mathrm{L}=−\mathrm{4}\:×\underset{{x}\rightarrow\pi/\mathrm{4}} {\mathrm{lim}}\frac{\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{cos}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)}\:=\:−\mathrm{4}\: \\ $$$$\:\mathrm{then}\:\frac{\mathrm{L}}{\mathrm{4}}\:=\:−\mathrm{1} \\ $$

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