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Question Number 84492 by john santu last updated on 13/Mar/20
lim_(x→(π/3)) ((sin (x−(π/3)))/(1−2cos (x))) =
$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({x}−\frac{\pi}{\mathrm{3}}\right)}{\mathrm{1}−\mathrm{2cos}\:\left({x}\right)}\:=\: \\ $$
Commented by john santu last updated on 13/Mar/20
lim_(x→0) ((sin (x))/(1−2cos (x+(π/3)))) = lim_(x→0) ((cos (x))/(2sin (x+(π/3)))) = (1/(2×((√3)/2))) = (1/( (√3)))
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({x}\right)}{\mathrm{1}−\mathrm{2cos}\:\left({x}+\frac{\pi}{\mathrm{3}}\right)}\:= \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cos}\:\left({x}\right)}{\mathrm{2sin}\:\left({x}+\frac{\pi}{\mathrm{3}}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$
Commented by john santu last updated on 13/Mar/20
other way sin (x−(π/3)) ≈ (x−(π/3)) cos (x) ≈ (1/2)−((√3)/2)(x−(π/3)) lim_(x→(π/3)) ((x−(π/3))/(1−2[(1/2)−((√3)/2)(x−(π/3))])) = lim_(x→(π/3)) (((x−(π/3)))/( (√3)(x−(π/3)))) = (1/( (√3))) .
$$\mathrm{other}\:\mathrm{way} \\ $$$$\mathrm{sin}\:\left(\mathrm{x}−\frac{\pi}{\mathrm{3}}\right)\:\approx\:\left(\mathrm{x}−\frac{\pi}{\mathrm{3}}\right) \\ $$$$\mathrm{cos}\:\left(\mathrm{x}\right)\:\approx\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{x}−\frac{\pi}{\mathrm{3}}\right) \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{{x}−\frac{\pi}{\mathrm{3}}}{\mathrm{1}−\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{x}−\frac{\pi}{\mathrm{3}}\right)\right]}\:= \\ $$$$\underset{{x}\rightarrow\frac{\pi}{\mathrm{3}}} {\mathrm{lim}}\:\frac{\left(\mathrm{x}−\frac{\pi}{\mathrm{3}}\right)}{\:\sqrt{\mathrm{3}}\left(\mathrm{x}−\frac{\pi}{\mathrm{3}}\right)}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:. \\ $$
Commented by mathmax by abdo last updated on 13/Mar/20
let l(x)=((sin(x−(π/3)))/(1−2cosx)) changement x−(π/3)=t give l(x)=A(t)=((sint)/(1−2cos(t+(π/3)))) =((sint)/(1−2(costcos(π/3)−sint sin((π/3))))) =((sint)/(1−2((1/2)cost−((√3)/2)sint))) =((sint)/(1−cost+(√3)sint)) x→(π/3) ⇔t→0 and A(t)∼ (t/((t^2 /2) +(√3)t)) =(1/((t/2)+(√3))) ⇒lim_(t→0) A(t)=(1/( (√3))) ⇒lim_(x→0) l(x)=(1/( (√3)))
$${let}\:{l}\left({x}\right)=\frac{{sin}\left({x}−\frac{\pi}{\mathrm{3}}\right)}{\mathrm{1}−\mathrm{2}{cosx}}\:\:{changement}\:{x}−\frac{\pi}{\mathrm{3}}={t}\:{give} \\ $$$${l}\left({x}\right)={A}\left({t}\right)=\frac{{sint}}{\mathrm{1}−\mathrm{2}{cos}\left({t}+\frac{\pi}{\mathrm{3}}\right)}\:=\frac{{sint}}{\mathrm{1}−\mathrm{2}\left({costcos}\frac{\pi}{\mathrm{3}}−{sint}\:{sin}\left(\frac{\pi}{\mathrm{3}}\right)\right)} \\ $$$$=\frac{{sint}}{\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}{cost}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sint}\right)}\:=\frac{{sint}}{\mathrm{1}−{cost}+\sqrt{\mathrm{3}}{sint}} \\ $$$${x}\rightarrow\frac{\pi}{\mathrm{3}}\:\Leftrightarrow{t}\rightarrow\mathrm{0}\:\:{and}\:\:\:{A}\left({t}\right)\sim\:\frac{{t}}{\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:+\sqrt{\mathrm{3}}{t}}\:=\frac{\mathrm{1}}{\frac{{t}}{\mathrm{2}}+\sqrt{\mathrm{3}}}\:\Rightarrow{lim}_{{t}\rightarrow\mathrm{0}} \:{A}\left({t}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}} \:\:{l}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$

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