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Question Number 107516 by bemath last updated on 11/Aug/20

⊚BeMath⊚ lim_(t→0) ((1−(√(cos 2t)))/(sin ((π/2)−t)−cos 2t)) ?

$$\:\:\:\:\circledcirc\mathcal{B}{e}\mathcal{M}{ath}\circledcirc \\ $$$$\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\sqrt{\mathrm{cos}\:\mathrm{2}{t}}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−{t}\right)−\mathrm{cos}\:\mathrm{2}{t}}\:? \\ $$

Answered by bemath last updated on 11/Aug/20

lim_(t→0) ((1−cos 2t)/(cos t−cos 2t)) × lim_(t→0) (1/(1+(√(cos 2t)))) = (1/2) × lim_(t→0) ((2sin^2 t)/(−2sin ((3/2)t)(sin −(t/2)))) =(1/2)×lim_(t→0) ((2sin^2 t)/(2sin (((3t)/2))sin ((t/2)))) =(1/2)×(4/3)=(2/3)

$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{2}{t}}{\mathrm{cos}\:{t}−\mathrm{cos}\:\mathrm{2}{t}}\:×\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{cos}\:\mathrm{2}{t}}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:×\:\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} {t}}{−\mathrm{2sin}\:\left(\frac{\mathrm{3}}{\mathrm{2}}{t}\right)\left(\mathrm{sin}\:−\frac{{t}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{2sin}\:^{\mathrm{2}} {t}}{\mathrm{2sin}\:\left(\frac{\mathrm{3}{t}}{\mathrm{2}}\right)\mathrm{sin}\:\left(\frac{{t}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Answered by Dwaipayan Shikari last updated on 11/Aug/20

lim_(t→0) ((1−(√(1−((4t^2 )/2))))/(cost−cos2t))=lim_(t→0) ((1−1+t^2 )/(1−(t^2 /2)−1+((4t^2 )/2)))=lim_(x→0) (t^2 /((3t^2 )/2))=(2/3)

$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\sqrt{\mathrm{1}−\frac{\mathrm{4}{t}^{\mathrm{2}} }{\mathrm{2}}}}{{cost}−{cos}\mathrm{2}{t}}=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1}+\frac{\mathrm{4}{t}^{\mathrm{2}} }{\mathrm{2}}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{t}^{\mathrm{2}} }{\frac{\mathrm{3}{t}^{\mathrm{2}} }{\mathrm{2}}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

Answered by mathmax by abdo last updated on 11/Aug/20

let f(t) =((1−(√(cos(2t))))/(sin((π/2)−t)−cos(2t))) we have cos(2t) ∼1−2t^2 ⇒(√(cos(2t)))∼1−t^2 ⇒1−(√(cos(2t)))∼t^2 also sin((π/2)−t) =cost ∼1−(t^2 /2) ⇒ sin((π/2)−t)−cos(2t) ∼ 1−(t^2 /2) −1+2t^2 =(3/2)t^2 ⇒f(t) ∼ (3/2) ⇒ lim_(t→0) f(t) =(3/2)

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{t}\right)\:=\frac{\mathrm{1}−\sqrt{\mathrm{cos}\left(\mathrm{2t}\right)}}{\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−\mathrm{t}\right)−\mathrm{cos}\left(\mathrm{2t}\right)}\:\mathrm{we}\:\mathrm{have}\:\mathrm{cos}\left(\mathrm{2t}\right)\:\sim\mathrm{1}−\mathrm{2t}^{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{cos}\left(\mathrm{2t}\right)}\sim\mathrm{1}−\mathrm{t}^{\mathrm{2}} \:\Rightarrow\mathrm{1}−\sqrt{\mathrm{cos}\left(\mathrm{2t}\right)}\sim\mathrm{t}^{\mathrm{2}} \:\:\mathrm{also}\:\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−\mathrm{t}\right)\:=\mathrm{cost}\:\sim\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−\mathrm{t}\right)−\mathrm{cos}\left(\mathrm{2t}\right)\:\sim\:\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}\:−\mathrm{1}+\mathrm{2t}^{\mathrm{2}} \:=\frac{\mathrm{3}}{\mathrm{2}}\mathrm{t}^{\mathrm{2}} \:\Rightarrow\mathrm{f}\left(\mathrm{t}\right)\:\sim\:\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{t}\right)\:=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by mathmax by abdo last updated on 11/Aug/20

sorry f(x)∼(2/3) ⇒lim_(t→0) f(x) =(2/3)

$$\mathrm{sorry}\:\mathrm{f}\left(\mathrm{x}\right)\sim\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow\mathrm{lim}_{\mathrm{t}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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