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Question Number 140574 by john_santu last updated on 09/May/21
lim_(x→(π/4)) ((1−tan^3 x)/(cos 2x)) =?
$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{3}} {x}}{\mathrm{cos}\:\mathrm{2}{x}}\:=? \\ $$
Answered by bramlexs22 last updated on 09/May/21
lim_(x→(π/4)) (((1+tan^2 x)(1−tan^3 x))/(1−tan^2 x)) = lim_(x→(π/4)) ((1+tan^2 x)/(1+tan x)) . lim_(x→(π/4)) (((1−tan x)(tan^2 x+tan x+1))/(1−tan x)) = (2/2) × (3/1) = 3.
$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\right)\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{3}} \mathrm{x}\right)}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{1}+\mathrm{tan}\:\mathrm{x}}\:.\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}−\mathrm{tan}\:\mathrm{x}\right)\left(\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}+\mathrm{tan}\:\mathrm{x}+\mathrm{1}\right)}{\mathrm{1}−\mathrm{tan}\:\mathrm{x}} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{2}}\:×\:\frac{\mathrm{3}}{\mathrm{1}}\:=\:\mathrm{3}. \\ $$
Answered by john_santu last updated on 09/May/21
lim_(x→(π/4)) ((cos^3 x−sin^3 x)/(cos^3 x(cos^2 x−sin^2 x))) = lim_(x→(π/4)) ((cos^2 x+cos x.sin x+sin^2 x)/(cos^3 x (cos x+sin x))) .lim_(x→(π/4)) ((cos x−sin x)/(cos x−sin x)) = ((3/2)/(((√2)/4)((√2)))) = 3.
$$\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:^{\mathrm{3}} {x}−\mathrm{sin}\:^{\mathrm{3}} {x}}{\mathrm{cos}\:^{\mathrm{3}} {x}\left(\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{2}} {x}\right)} \\ $$$$=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:{x}.\mathrm{sin}\:{x}+\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:^{\mathrm{3}} {x}\:\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)}\:.\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\frac{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}−\mathrm{sin}\:{x}} \\ $$$$=\:\frac{\frac{\mathrm{3}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\sqrt{\mathrm{2}}\right)}\:=\:\mathrm{3}.\: \\ $$
Answered by EDWIN88 last updated on 09/May/21
x=(π/4)+z ⇒lim_(z→0) ((1−tan^3 (z+(π/4)))/(cos (2z+(π/2)))) = lim_(z→0) (((tan^2 (z+(π/4))+tan (z+(π/4))+1)(1−tan (z+(π/4))))/(−sin 2z)) = 3 × lim_(z→0) ((cos (z+(π/4))−sin (z+(π/4)))/(−cos (z+(π/4)) sin 2z)) = −(6/( (√2))) × lim_(z→0) ((−sin (z+(π/4))−cos (z+(π/4)))/(2cos 2z)) = −(6/( (√2))) ×((−(√2))/2) = 3
$$\mathrm{x}=\frac{\pi}{\mathrm{4}}+\mathrm{z}\:\Rightarrow\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{3}} \left(\mathrm{z}+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{cos}\:\left(\mathrm{2z}+\frac{\pi}{\mathrm{2}}\right)} \\ $$$$=\:\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{tan}\:^{\mathrm{2}} \left(\mathrm{z}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{tan}\:\left(\mathrm{z}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{1}\right)\left(\mathrm{1}−\mathrm{tan}\:\left(\mathrm{z}+\frac{\pi}{\mathrm{4}}\right)\right)}{−\mathrm{sin}\:\mathrm{2z}} \\ $$$$=\:\mathrm{3}\:×\:\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left(\mathrm{z}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{sin}\:\left(\mathrm{z}+\frac{\pi}{\mathrm{4}}\right)}{−\mathrm{cos}\:\left(\mathrm{z}+\frac{\pi}{\mathrm{4}}\right)\:\mathrm{sin}\:\mathrm{2z}} \\ $$$$=\:−\frac{\mathrm{6}}{\:\sqrt{\mathrm{2}}}\:×\:\underset{\mathrm{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{−\mathrm{sin}\:\left(\mathrm{z}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{cos}\:\left(\mathrm{z}+\frac{\pi}{\mathrm{4}}\right)}{\mathrm{2cos}\:\mathrm{2z}} \\ $$$$=\:−\frac{\mathrm{6}}{\:\sqrt{\mathrm{2}}}\:×\frac{−\sqrt{\mathrm{2}}}{\mathrm{2}}\:=\:\mathrm{3} \\ $$

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