Question-159613

Question Number 159613 by cortano last updated on 19/Nov/21

Commented by tounghoungko last updated on 19/Nov/21

A = lim_(x→0) ((2sin x(1−cos x))/(sin x(1−cos^3 x))) A= lim_(x→0) ((2(1−cos x))/((1−cos x)(cos^2 x+cos x+1))) A= (2/3)

$$\:{A}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:{x}\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{\mathrm{sin}\:{x}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{3}} {x}\right)} \\ $$$$\:{A}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2}\left(\mathrm{1}−\mathrm{cos}\:{x}\right)}{\left(\mathrm{1}−\mathrm{cos}\:{x}\right)\left(\mathrm{cos}\:^{\mathrm{2}} {x}+\mathrm{cos}\:{x}+\mathrm{1}\right)} \\ $$$$\:{A}=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Commented by tounghoungko last updated on 19/Nov/21

D= lim_(x→(π/6)) (((2sin x−1)(sin x−1))/(cos x(1−2sin x))) D= (((1/2)−1)/(−((√3)/2))) = (1/2) ×(2/( (√3))) =(1/( (√3)))

$$\:{D}=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\left(\mathrm{2sin}\:{x}−\mathrm{1}\right)\left(\mathrm{sin}\:{x}−\mathrm{1}\right)}{\mathrm{cos}\:{x}\left(\mathrm{1}−\mathrm{2sin}\:{x}\right)} \\ $$$$\:{D}=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}}{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:×\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\: \\ $$

Commented by tounghoungko last updated on 19/Nov/21

E= lim_(x→(π/3)) ((2cos x−1)/(3−4(1−cos^2 x))) E= lim_(x→(π/3)) ((2cos x−1)/((2cos x−1)(2cos x+1))) E= (1/(2((1/2))+1)) = (1/2)

$$\:{E}=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:{x}−\mathrm{1}}{\mathrm{3}−\mathrm{4}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)} \\ $$$$\:{E}=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{2cos}\:{x}−\mathrm{1}}{\left(\mathrm{2cos}\:{x}−\mathrm{1}\right)\left(\mathrm{2cos}\:{x}+\mathrm{1}\right)} \\ $$$$\:{E}=\:\frac{\mathrm{1}}{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by tounghoungko last updated on 19/Nov/21

C=lim_(x→(π/4)) (((cos x−sin x)(1+(1/2)sin 2x))/(cos^3 x (cos x−sin x))) C= ((1+(1/2))/((((√2)/2))^3 )) = (3/2) ×(8/(2(√2))) = (6/( (√2))) = 3(√2)

$$\:{C}=\underset{{x}\rightarrow\frac{\pi}{\mathrm{4}}} {\mathrm{lim}}\:\frac{\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\right)}{\mathrm{cos}\:^{\mathrm{3}} {x}\:\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right)} \\ $$$$\:{C}=\:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}{\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{3}} }\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:×\frac{\mathrm{8}}{\mathrm{2}\sqrt{\mathrm{2}}}\:=\:\frac{\mathrm{6}}{\:\sqrt{\mathrm{2}}}\:=\:\mathrm{3}\sqrt{\mathrm{2}} \\ $$

Commented by tounghoungko last updated on 19/Nov/21

B= lim_(x→(π/6)) (((2sin x−1)(sin x−1))/(2sin x−1)) B=(1/2)−1=−(1/2)

$$\:{B}=\:\underset{{x}\rightarrow\frac{\pi}{\mathrm{6}}} {\mathrm{lim}}\:\frac{\left(\mathrm{2sin}\:{x}−\mathrm{1}\right)\left(\mathrm{sin}\:{x}−\mathrm{1}\right)}{\mathrm{2sin}\:{x}−\mathrm{1}} \\ $$$$\:{B}=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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